Maths and Chemistry chapter

MATHS AND

CHEMISTY

THIS EBOOK WAS PREPARED

AS A PART OF THE COMENIUS PROJECT

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by the students and the teachers from:

BERKENBOOM HUMANIORA BOVENBOUW, IN SINT-NIKLAAS ( BELGIUM)

EUREKA SECONDARY SCHOOL IN KELLS (IRELAND)

LICEO CLASSICO STATALE CRISTOFORO COLOMBO IN GENOA (ITALY)

GIMNAZJUM IM. ANNY WAZÓWNY IN GOLUB-DOBRZYŃ (POLAND)

ESCOLA SECUNDARIA COM 3.º CICLO D. MANUEL I IN BEJA (PORTUGAL)

IES ÁLVAREZ CUBERO IN PRIEGO DE CÓRDOBA (SPAIN)

This project has been funded with support from the European Commission.

This publication reflects the views only of the author, and the

Commission cannot be held responsible for any use which may be made of the

information contained therein.

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Math is everywhere. It is an universal language which everyone needs. We use math

every day, everywhere and anytime. We use it automatically. It can be applied in simple tasks,

like figuring out how many time you have until your next class, or in long and complicated

tasks, for example doing your tasks.

Chemists use math for a variety of tasks. We balance the equation of a chemical

reaction, use mathematical calculations that are absolutely necessary to explore important

concepts in chemistry, and utilize dimensional analysis to find any range of information

about reactions from finding the mass of chemicals reacted to the concentration of a chemical

in a solution. Math is also used to calculate energy in reactions, compression of a gas, grams

needed to add to a solution to reach desired concentration, and quantities of reactants needed

to reach a desired product. It is important to know how to mathematically handle chemistry

problems in order to understand what they mean and how to prepare specific quantities of

chemicals.

Math applied to Chemistry

Overall and simpler operations

Math and chemistry are closely linked. We practically need mathematical operations to

do everything in chemistry. Sum, subtraction, division and multiplication are essential things

we need to know before starting to think about being a chemist. Other operations, like the

simple rule of three and proportions, are basic tools you need to know before starting to study

chemistry. Have no doubt that those are key components to have success in chemistry.

Ex:

If there’s 1 mol of Cl in one molecule of NaCl how many moles are in 5 molecules of NaCl

1 1

5 x

x = 5 mol

We used the simple rule of three to figure out the chemical quantity of sodium in 5

molecules of sodium chlorite NaCl.

Take chemistry out of the equation

Equations are perhaps the most important math principal you need to learn. We use all

kinds of equations in chemistry.

Ex: N = n x NA

A = Z + N

Eradiation = Eremoval + Ekinetics

If you need to succeed in chemistry you need to know how to make all kinds of

equations.

Different sizes, different views

Chemists transform very often some units of measure into others, according to the

variables of the experiment. To do this they need math. For example, to transform meters in

light-years we use this formula:

1 l.y. = 9.47 ·1015 m

Another example is changing Celsius to Kelvin, two units to measure temperature.

T(K) = T(ºC) + 273

Measuring

To measure a certain object you need math. If you measure it directly, which means

you are there measuring at the spot, you need to know the units therefore you need math. If

you are measuring it indirectly, through an equation or an expression, you also need math

because you need to know how to solve them. Here are some examples:

Directly:

What is the length of the book?

To measure the length of the book you need to know which unit is the ruler. Of course

it is probably cm and you know this without thinking, but that happens because you learn it

in math class.

Indirectly:

A certain object has a weight of 210 g and a volume of 20 mL. Calculate the density of

the object.

Solution: p = m : V (=)

(=) p = 10,5 g/mL

In this case, we are using an equation to measure a chemical greatness (density).

Scientific Notation

There are certain greatness’s used in chemistry that are too big or too small to

represent the full number. So in chemistry is frequently used something we learn in math,

scientific notation. For example, to represent the number of meters in one parsec we use this

number 3.09·1016m because it’s too big. However to represent mercury’s ionization energy we

use this number 1,21·10-18 J/e.

Also, operations with numbers in scientific notation is commonly used in chemistry-

Ex: In the hydrogen atom, an electron “jumps” from the first level of energy to the

second one. Determine the energy of the photon necessary for this transition to happen.

Resolution: Ef = E2 – E1

Avogadro’s number

Fractions

Scientific notation is not the only form of number representation used in chemistry.

Fractions were invented by mathematicians and now is commonly used in chemistry to

represent exact numbers.

Ex: 1:3 = 0.(3)

Statistics

Sometimes it’s used in chemistry charts, tables, graphics, and averages, among others.

Are often used to organize a very large group or to determine or register something’s

behavior. A much known example of the use of statistic in chemistry is the Periodic Table.

The Periodic Table is a table (concept taken from math) that contains every known chemical

element in the world, organized by their atomic number and with some of the atom’s

properties. In this case chemists used statistic to have an organized and easy way to access

information. Another common use for statistics is graphics. Chemists regularly use graphics

to predict or register something’s behavior. For example, it is used a line graphic to register

the several ionization energies of the first 20 elements of the Periodic Table.

We can use line graphics but we can also use other types. It’s frequently used bar

graphs to represent and understand better the evolution of the Earth’s atmosphere.

Diagrams is another concept that is very used in chemistry, especially when trying to

explain an argument. It is sometimes easier to understand when is summarized in a diagram.

These are the forms of data representation. However, are used calculus learnt in

statistics. A common example is average. In chemistry is used to calculate the relative isotopic

mass or to compare results.

In conclusion, knowing statistics is very important in chemistry.

In conclusion

To study chemistry, we need math. It is inevitable. Math is an essential part of the

human knowledge, we can’t live without it.

You can also watch the presentation prepared by the Portuguese student here: LINK

http://prezi.com/fdljxuzjemhw/?utm_campaign=share&utm_medium=copy&rc=ex0share

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Balancing chemical equations

Stoichiometry - is a branch of chemistry that deals with the relative quantities of reactants

and products in chemical reactions. Stoichiometry is the mathematics behind the science of

chemistry.

A chemical equation is an easy way to represent a chemical reaction—it shows which

elements react together and what the resulting products will be. By the Law of Conservation

of Mass, the number of atoms must be the same on both sides because these atoms cannot be

created or destroyed in a reaction.

The number of atoms that we start with at the beginning of the reaction must equal the

number of atoms that you end up with.

When the number of atoms of reactants matches the number of atoms of products, then the

chemical equation is said to be balanced.

We would like to present a simple method of defining the coefficients in the equations of

chemical reactions with the help of a system of linear algebraic equations that describes the

material balance in a chemical reaction.

Example 1:

CH4 + O2 → CO2 + H2O

I. First we use each element to produce an equation involving the

coefficient letters. We need to find the smallest possible positive

integers a, b, c, and d such that the following chemical equation.

aCH4 + bO2 → cCO2 + dH2O

is balanced.

Carbon: a = c There is 1 carbon atom in the a term and 1 in the c term.

Hydrogen: 4a = 2d There are 4 hydrogen atoms in the a term and 2 in the d term.

Oxygen: 2b = 2c + d There are 2 oxygen atoms in the b term and 2 in the c term

and 1 in the d term.

II. As a result we get a system of three linear equations with four unknowns:

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III. This system has an infinite number of solutions, but we have to get the minimal

natural values only. Since we would like the smallest integral solutions, = 1 works

well.

The system has the following solution - the coefficients are = 1, = 2, = 1 and = 2

When writing our balanced equation, the coefficient 1 is assumed and can be omitted,

yielding the formula:

CH4 + 2O2 → CO2 + 2H2O

The balanced chemical equation has one mole of methane reacting with two moles of oxygen

gas to form one mole of carbon dioxide and two moles of water.

Example 2:

Plants use the process of photosynthesis to convert carbon

dioxide and water into glucose and oxygen. This process helps

remove carbon dioxide from the atmosphere. Balance the

following equation for the production of glucose and oxygen

from carbon dioxide and water.

CO2 + H2O → C6H12O6 + O2

This equation needs to be balanced. We must find coefficients

a, b, c and d in the reactants and products - rewrite the

equation as:

aCO2 + bH2O → cC6H12O6 + dO2

where the numbers of carbon, hydrogen, and oxygen atoms are the same on both sides of the

equation:

Carbon: a = 6c

Oxygen: 2a + b = 6c + 2d

Hydrogen: 2b = 12c

Since we would like the smallest integral solutions, = 6 works well and the coefficients are

= 6, = 6, = 1 and = 6

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Examples of Balancing Chemical Equations

Consider the unbalanced reactions.

1. FeCl3 + NH4OH → Fe(OH)3 + NH4Cl

Assign each molecule a variable a, b, c, d since we have 4 expressions in the reaction

We need to find the smallest possible positive integers a, b, c, and d such that the following

chemical equation.

aFeCl3 + bNH4OH → cFe(OH)3 + dNH4Cl

We can obtain a set of linear equations in these variables by considering the number of times

each type of atom occurs on each side of this equation

1) Fe: a = c There is 1 iron atom in the a term and 1 in the c term.

Cl: 3a = d There are 3 chlorine atoms in the a term and 1 in the d term

N: b = d There is 1 nitrogen atom in the b term and d in the c term

H: 5b = 3c + 4d There are 5 hydrogen atoms in the b term and 3 in the c term

and 4 in d term.

O: b = 3c There is 1 oxygen atom in the a term and 1 in the c term

2) As a result we will get a system of five linear equations with four unknowns:

3)

4)

FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl

2. aNaCl + bSO2 + cH2O + dO2 → eNa2SO4 + f HCl

From this equation we obtain the following relations in the unknowns a, b, c, d, e, and f:

1) Na: a = 2e

Cl: a = f

S: b = e

O: 2b + c + 2d = 4e

H: 2c = f

2)

3)

4)

To calculate the smallest possible positive integer value of a, we have to find the least

common denominator of b, c, d, and e which in this case is 4. If we work out the above

equations we calculate the values of our 5 variables to be:

For a = 4 we have:

b = 2

c = 2

d = 1

e = 2

f = 4

4NaCl + 2SO2 + 2H20 + 02 → 2Na2SO4 + 4 HCl

3. aCaCO3 + bHNO3 → cCa(NO3)2 + dH2O + eCO2

Specifying the values a, b, c, d and e for the coefficients of this equation we have:

1) Ca: a = c

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www.chemistrysmostwanted.wikispaces.com

C: a = e

H: b = 2d = 2a

N: b = 2c = 2a

O: 3a + 3b = 6c + d + 2e

Solving simultaneously and using the smallest integers we have

1)

2) For a = 1

b = 2a = 2

c = a = 1

d = a = 1

e = a = 1

CaCO3 + 2 HNO3 → Ca(NO3)2 + H2O + CO2

How is marble eroded by acid rain? Atmospheric sulfur dioxide

combines with rainwater to create sulfurous acid. The primary

component of marble is calcium carbonate (CaCO3). The sulfurous

acid reacts with the CaCO3 in the marble and dissolves it. Marble

statues (CaCO3) attacked by acid rain (containing HNO3).

4. aHCl + bK2CO3 → cCO2 + dH2O + eKCl The equation for each atom looks like:

1) H: a = 2d d =

Cl: a = e

K: 2b = e b =

C: b = c c =

O: 3b = 2c + d

So we have now after some canceling:

2) b =

c =

d =

e =

For a = 2 we have:

3) b = 1 c = 1

d = 1

e = 2

2HCl + K2CO3 → CO2 + H2O + 2KCl

How to calculate density

Density is a characteristic property of a substance. The density of

a substance is the relationship between the mass of the substance and how

much space it takes up (volume). The density, or more precisely, the

volumetric mass density, of a substance is its mass per unit volume.

Mathematically, density is defined as mass divided by volume :

v

m

volume

massd

To calculate the specific gravity (S.G.) of an object, you compare the object's

density to the density of water.

Examples of densities:

Solids: Liquids: Gases:

Silver = 10.49 Milk = 1.020-1.050 Air = 0.001293

Aluminum = 2.7 Glucose = 1.350-1.440 Argon = 0.001784

Diamond = 3.01-3.25 Glycerine = 1.259 Chlorine= 0.0032

Gold = 19.3 Flourine = 0.001696

Magnesium = 1.7 Helium = 0.000178

Platinum = 21.4 Neon = 0.0008999

Example 1

Calculate the density of the cube made of silver, whose weight is 262.5 grams and the volume

is 25 cm ³.

Given: Calculate

m = 262.5 g d = ?

V = 25 cm³

1 d =

Answer: Density of silver is 10.5

.

Example 2

There is 250 cm ³ of ethyl alcohol in the glass vessel. The volume of alcohol is 0.197 kg

Calculate the density of alcohol and enter the result in grams per cubic centimeter.

Given: Calculate:

m = 0.197 kg d = ?

V = 250 cm³

1 m = 0.197kg= 0.197 g

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2 d =

The density of ethyl alcohol is 0.79

Example 3

Calculate the mass of 300 cm³ gasoline which density is 0.75

.

Given: Calculate:

V = 300 cm³ m = ?

d = 0.75

1 d =

m = d V

2 m = 0.75

300cm³ = 225 g

Answer: The mass of gasoline is 225g.

Example 4

A container of volume 0.05m3 is full of ice. When the ice melts into water, how many kg of

water should be added to fill it up? (density of ice = 900

; density of water = 1000

)

Given: Calculate:

dice = 900

mwater = ?

dwater = 1000

Vice = 0.05 m³

1 mice = d V

mice =

2

We should add 5 kg of water to fill the container up.

Example 5

A rubber ball has a radius of 2.5 cm. The density of rubber is 1.2

. What is the mass of the

ball?

Given: Calculate:

r = 2.5 cm m = ?

d = 1.2

1 First we calculate the volume of a ball:

2

Answer: The mass of the ball is about 78.5 g.

Example 6

A 5.6-gram marble put in a graduated cylinder raises the water from 30 mL to 32 mL. What is

the marble’s density?

Given: Calculate:

m = 6g d = ?

1 First we calculate the volume of marble:

2

d =

The density of marble is

.

Example 7

A small rectangular slab of lithium has the dimensions 20.9 mm by 11.1 mm by 11.9 mm. Its

mass is 1.49·103 mg. What is the density of lithium in

?

Given: Calculate:

a = 20.9 mm = 2.09 cm d = ?

b = 11.1 mm = 1.11 cm

c = 11.9 mm = 1.19 cm

m = 1.49·103 mg = 1490 mg = 1.49 g

1

V

2

d =

The density of lithium is 0.53

.

Example 8

Find the mass of air inside a room measuring 10m×8m×3m, if the density of air is 1.28

.

Given: Calculate:

a = 10m m = ?

b = 8m

c = 3m

d = 1.28

1

V

2

m = d V

m =

The mass of the air inside the room is

2mL

Picture from:www.monarchbearing.com

Example 9

You have two stainless steel balls. The larger has a mass of 25 grams

and a volume of 3.2cm3. The smaller has a mass of 10 grams.

Calculate the volume of the smaller ball.

Given: Calculate:

m1 = 25g V2 = ?

V1 = 3.2cm3

m2 = 10g

1

dsteel =

2

V2 =

=

The volume of the smaller ball is

Since density is a characteristic property of a substance, each liquid has its

own characteristic density. The density of a liquid determines whether it

will float on or sink in another liquid. A liquid will float if it is less dense

than the liquid it is placed in. A liquid will sink if it is more dense than the

liquid it is placed in.

Example 10

A rectangular object is 10 centimeters long, 5 centimeters high, and 20 centimeters wide. Its

mass is 800 grams. Will the object float or sink in water? Remember that the density of water

is about 1

.

Given: Calculate:

a = 10cm d = ?

b = 5cm

c = 20cm

m = 800g

1

V

2 d =

The object will float.

carbon oxygen

carbon hydrogen oxygen

Picture from: www.annekeckler.com

Percentage composition

Percentage composition is just a way to describe what

proportions of the different elements there are in a

compound.

If you have the formula of a compound, you should be

able to work out the percentage by mass of an element

in it.

%Composition A=

Example 1

What is the percentage composition of carbon and oxygen in

?

First we need to find the mass of the compound.

Molar mass of compound: 12.01+ 2

Next we need to find the mass of carbon and oxygen in the

compound.

Molar mass of carbon: 12.01

Molar mass of oxygen: 32

Then we should divide the mass of each element by the mass of

the compound and multiply by 100%.

The percentage composition of carbon is:

%C =

The percentage composition of oxygen is:

% O =

72.71%

Example 2

What is the percentage composition by mass of the elements in

the compound

We start by finding the atomic weights.

Molar mass of C: = 12.01

Molar mass of H: = 1.01

Molar mass of O: = 16.00

Work out the molecular weight of glucose: 6 12.01 + 12 1.01 + 6 16 = 180

potassium chromium oxygen

Picture from: www.commons.wikimedia.org

Picture from: www.aromaticscience.com

Mass of carbon: 6 12.01 = 72.06

Mass of hydrogen: 12 1.01 = 12.12

Mass of oxygen: 6 16 = 96


The percentage composition of hydrogen is:


Examples 3

What is the percentage composition of chromium in ?

Molar mass of compound: 2 39 + 2 52 + 7 16 = 249

Molar mass of chromium:

The percentage composition of Cr is:

Examples 4

What is the percentage by mass of nitrogen in ammonium nitrate, NH4NO3 an important

source of fertilizer?

Molar mass of N: = 14.01

Molar mass of compound: 2 + 4 + 3 16 =

The percentage composition of nitrogen is:

The percentage composition of N is 35%

Examples 5

Cinnamaldehyde, C9H8O, is responsible for the characteristic

odour of cinnamon. Determine the percentage composition of

cinnamaldehyde by calculating the mass percents of carbon,

hydrogen, and oxygen.

The molecular formula of cinnamaldehyde is C9H8O.

Molar mass of C: = 12.01

Molar mass of H: = 1.01

Molar mass of O: = 16.00

First we calculate the molar mass of cinnamaldehyde:

12.01 + 8 1.01 + 16 = 132.17

Mass of carbon: = 9 12.01 = 108.09

Mass of hydrogen: = 8 1.01 = 8.08

Mass of oxygen: = 16


The percentage composition of hydrogen is:


Determining empirical and molecular formulas

The empirical formula is the simplest formula for a compound. A molecular formula is the

same as or a multiple of the empirical formula, and is based on the actual number of atoms of

each type in the compound. For example, if the empirical formula of a compound is C3H8 , its

molecular formula may be C3H8 , C6H16 , etc.

An empirical formula is often calculated from elemental composition data. The weight

percentage of each of the elements present in the compound is given by this elemental

composition.

Using basic mathematics skills like ratio, percentage, linear equations and system of linear

equations we can determine the empirical formula of an unknown compound from its atomic

masses and percent composition.

Example 1

Analysis of a compound gives 30.43 % N and 69.57% O. The mass for this compound is 92u.

What is its molecular formula?

Given Find

Formula:

%N = 30.43%

%O = 69.57%

Molar mass of N: = 14

Molar mass of O: = 16

First method

We assume that the molecular weight is 100% and calculate the mass of nitrogen

It means that 28 in the compound is for the atoms of nitrogen and the rest:

92 - 28 = 64 is for the oxygen.

The numbers of atoms in the compound is:

atoms of N

atoms of O

The formula is .

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Second method

In the second method we use the system of linear equations.

First equation: Second equation:

The formula molecular is .

Empirical Formula Calculation Steps

Step 1

If we have masses go onto step 2.

If we have %. Assume the mass to be 100g, so the % becomes grams e.g. 40% of a compound

is carbon. 40% of 100 g is 40 grams.

Step 2

Determine the moles of each element.

Step 3

Determine the mole ratio by dividing each elements number of moles by the smallest value

from step 2.

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Step 4

Round your ratio to the nearest whole number as long as it is “close.” For example, 1.99987

can be rounded to 2, but 1.3333 cannot be rounded to 1. It is four-thirds, so we must multiply

all ratios by 3 to rid ourselves of the fraction. If we have the empirical formula C1.5H3O1 we

should convert all subscripts to whole numbers, multiply each subscript by 2. This gives us

the empirical formula C3H6O2. Thus, a ratio that involves a decimal ending in .5 must be

doubled. We should double, triple to get an integer if they are not all whole numbers.

Example 2

A sulfide of iron was formed by combining 2.233 g of Fe with 1.926 g of S calculate the

empirical formula.

Molar mass of iron: = 55.85

Molar mass of sulfur: = 32.1

1 Mass of iron: = 2.233g

Mass of sulfur: = 1.926g

2 Convert masses to amounts in moles

Numbers of moles of iron:

Numbers of moles of sulfur:

3 Divide these numbers of moles by the smallest number (0.03998 in this case)

Fe ⇒

S ⇒

Preliminary formula is:

Now we should multiply to get a whole number. In order to turn 1.5 into a whole number, we

need to multiply by 2 – therefore all results must be multiplied by 2.

The simplest formula is:

Example 3

Find the empirical formula for a compound containing 36.5% sodium, 25.4% sulfur and

38.1% oxygen.

Molar mass of sodium Na: = 23

Molar mass of sulfur S: = 32.1

Molar mass of oxygen O: = 16

·2

1

We assume that we have 100g of total material, and % becomes grams

Mass of sodium: = 36.5g

Mass of hydrogen: = 25.4g

Mass of oxygen: = 38.1g


Numbers of moles of sodium:


Numbers of moles of oxygen:


Na ⇒

S ⇒

O ⇒

The formula is sodium sulfite.

Example 4

The composition of ascorbic acid (vitamin C) is 40.92% carbon, 4.58%

hydrogen, and 54.50% oxygen. What is the empirical formula for vitamin C?

Molar mass of carbon C: = 12.01

Molar mass of hydrogen H: = 1.01


1

We are only given mass %, and no weight of the compound so we will assume 100g of the

compound, and % becomes grams

Mass carbon = 40.92g

Mass hydrogen = 4.58g

Mass oxygen = 54.5g

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numbers of moles of carbon:

numbers of moles of hydrogen:

numbers of moles of oxygen:


C ⇒

H ⇒ 4.533.4=1.33 O ⇒ 3.43.4=1

Preliminary formula is:

Multiply to get a whole number. In order to

turn 1.33 into a whole number, we need to

multiply by 3 – therefore all results must be

multiplied by 3

The simplest empirical formula of vitamin C is

C3H4O3

Example 5

Muscle soreness from physical activity is caused by a buildup of

lactic acid in muscle tissue. Analysis of lactic acid reveals it to be

40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar

mass is 90.088

. What are the empirical and molecular formulas?




1

We are only given mass %, and no weight of the compound so we will assume 100g of the

compound, and % becomes grams

Mass carbon = 40g


Mass oxygen = 53.3g


numbers of moles of carbon:

·3

numbers of moles of hydrogen:

numbers of moles of oxygen:

3 Divide these numbers of moles by the smallest number

C ⇒

H ⇒

O ⇒

The empirical formula of lactic acid C is C1H2O1

4 Next we calculate the empirical formula weight: 12.01 + 21.01+ 16 = 30.03

5 Divide the molecule weight 90.08

by the empirical formula weight

The molecular formula of lactic acid is C3H6O3.

Example 6

A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by weight. The molecular

weight for this compound is 64.07

. What is its molecular formula?

Molar mass of sulfur S: = 32.1


1 First we will find the empirical formula. We assume 100 g of the compound is present and

change the percents to grams:

Mass sulfur = 50.05g

Mass oxygen = 49.95g

2 Then we convert the masses to moles:


Numbers of moles of oxygen:

3 Divide by the lowest, seeking the smallest whole-number ratio:

S ⇒

O ⇒

4 And now we can write the empirical formula: SO2

5 Next we calculate the empirical formula weight

32 + 216 = 64

6 Divide the molecule weight by the empirical formula weight

7 Use the scaling factor computed just above to determine the molecular formula:

SO2 times 1 gives SO2 for the molecular formula.

Example 7

An analysis of nicotine, an addictive compound found in tobacco leaves, shows that it is

74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol. What are its empirical and

molecular formulas?



Molar mass of nitrogen N: = 14

1

We assume 100g of the compound, and % becomes grams

Mass carbon = 74g


Mass nitrogen = 17.35g


Numbers of moles of carbon:

Numbers of moles of hydrogen:

Numbers of moles of nitrogen:

3 Divide by the lowest, seeking the smallest whole-number ratio:

C ⇒

O ⇒

N ⇒

4 And now we can write the empirical formula:

5 Next we calculate the empirical formula weight

5·12.01 + 71.01+ 14 = 81.12

Picture from:www.previewcf.turbosquid.com

6 Divide the molecule weight by the empirical formula weight

The molecular formula of nicotine is .

Concentration by percent

Medicated syrup is an example of

the concentration.

Brine and syrup

Brine is a concentrated solution of sodium chloride in

water.

In Poland we use brine to prepare our special

cucumbers - gherkins.

Syrup is a concentrated solution of sugar in water.

Example of syrup is fruit syrup

Calculating concentration of a chemical solution requires basic math skills like knowing percentage,

equations or system of equations.

Solutions are homogeneous mixtures of solute and solvent.

Solvent - the most abundant substance in a solution. In a liquid solution, the solvent does the

dissolving.

Solute - the other substance in a solution. In a liquid solution, the solute is dissolved.

Concentration refers to the amount of solute that is dissolved in a solvent. The concentration

of a solution in percent can be expressed in two ways: as the ratio of the volume of the solute

to the volume of the solution or as the ratio of the mass of the solute to the mass of the

solution.

To calculate percent concentration we use a formula:

The mass percent of a solution is a way of expressing its concentration. Mass percent is found by

dividing the mass of the solute by the mass of the solution and multiplying by 100; e.g. a solution of

NaOH that is 28% NaOH by mass contains 28 g of NaOH for each 100 g of solution.

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Example 1:

Calculate concentration of sugar in compote. For every kilogram of fruits

you need syrup made from 0.4 liter of water and 0.4 kilogram of sugar.

Given Calculate

mwater = 0.4L= 400g Cp = ?

mfruit = 0.4kg= 400g

msyrup = 400g+ 400g= 800g

Answer: Concentration of syrup is 50%.

Example 2:

We mingled 200L milk which contain 2% butterfat and 50L milk which contain 4% butterfat.

Calculate final percent concentration in milk.

Given: Calculate

Cp = ?

1

2

Answer: We got 2.4% milk.

Example 3

A bottle of the antiseptic hydrogen peroxide H2O2 is labeled 3%. How

many mL H2O2 are in a 473 mL bottle of this solution?

Given: Calculate

Cp = 3%

1

Answer: There are about 14.2mL of H2O2 in a bottle of this solution.

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Example 4

How many grams of salt do you need to make 500

grams of a solution with a concentration of 5% salt?

Given Calculate

Cp = 5% msolute = x ?

msolution =500g

msolute = x = 25g

Answer: We need 25g of salt.

Example 5

How many grams of water must be evaporated from 10 grams of a 40% saline solution to

produce a 50% saline solution?

Given Calculate

x the amount of evaporated water

( in grams)

The amount of salt in the beginning and after the evaporation of water is the same,

Answer: 2 grams of water must be evaporated from the 40% saline solution.

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In the beginning In the end

10g of solution

40%·10 the

amount of salt

(10-x) of solution

50%·(10-x) the

amount of salt

x g of water

Example 6

How many grams of salt must be added to 30kilos of a 10% salt solution to increase the salt

concentration to 25%. How many kilos of salt were added?

Given Calculate

x the amount of added salt ( in kilos)

The amount of water in the beginning and after adding salt is the same.

Answer: 0.6 kilograms of salt must be added to the solution to increase the salt

concentration.

To solve word problems involving percent concentration amounts, knowledge of solving

systems of equations and percents is necessary. Below there are some percent

concentration problems involve solving systems of equations when mixing two liquids with

differing percent concentration amounts.

Example 7

A 16% salt solution is mixed with a 4% salt solution. How many milliliters of each solution are

needed to obtain 600 milliliters of a 10% solution?

The table below help us organise information.

Amount of

solution

(in mL)

Percent Total

16% salt solution x 0.16 0.16x

4% salt solution y 0.4 0.4y

mixture x + y = 600 0.10 0.10·600 = 60

The liters of salt solution from

the 16% solution, plus the liters

of acid in the 30% solution, add

up to the liters of acid in the

10% solution

The first column is for

the amount of each

item we have.

In the beginning In the end

30kg of solution

10%·30 the

amount of salt

90%·30 the

amount of water

(30+x) of solution

25%·(30+x) the

amount of salt

75%·(30+x) the

amount of salt

x g of salt

Answer: 300mL of these solutions are needed to obtain 600 milliliters of a 10% solution.

Example 8

How much 10% sulfuric acid (H2SO4) must be mixed with how

much 30% sulfuric acid to make 200 milliliters of 15% sulfuric

acid?

Let's organise the information in the table

Volume Percent Amount of salt

10% sulfuric acid x 0.10 0.1x

30% sulfuric acid y 0.30 0.3y

mixture x + y = 200 0.15 0.15·200 = 30

Now that the table is filled, we can use it to get two equations. The "volume" and "amount of

acid" columns will let us get two equations.

Since x + y = 600, then x = 600 – y.

We can substitute for x in our second equation, and eliminate

one of the variables

Since x + y = 200, then x = 200 – y.

The liters of sulfuric acid from

the 10% solution, plus the

liters of acid in the 30%

solution, add up to the liters of

acid in the 15% solution

The first column is for

the amount of each

item we have.

Answer: 150mL 10% sulfuric acid must be mixed with 50mL 30% sulfuric acid to make 200

milliliters of 15% sulfuric acid.

We can substitute for x in our second equation, and eliminate

one of the variables

PH AND LOGARITHMS

WHAT IS pH?

SOME INFORMATION ABOUT pH

PH is the scale of the measure of acidity and basicity of a liquid

solution, based on the concentration of positive ions and negative

ions.

The term "PH" was introduced in 1909 by the Danish chemist

Soren Sorensen.

The PH's scale goes from 0 to 14. 0

stands for the maximum acidity (for

example hydrochloric acid), 14 stands

for the maximum basicity (so it is

sodium hydroxide). The medium

value is 7 and it belongs to distilled

water at the temperature of 25°C and

stands for a neutral solution.

The traditional indicator for PH is the

litmus test, a special paper that turns

from green to red immersed in an acid

solution. At the contrary in a basic

solution the paper turns from green to

dark blue.

WHAT IS pOH?

The pOH scale calculates the concentration of OH ions in a liquid solution. It is the exact

opposite of the pH and they are complementary. The logarithmic function of one completes

the one of the other too.

WHAT IS A LOGARITHM?

It defines logarithm in basis a of a number N the exponent it has to give to a to get the

number N (N is called the argument of the logarithm).

CONNECTION BETWEEN LOGARITHMS AND PH

We have already given a definition of the pH, but it can be given another definition linked to

the logarithms. In fact the pH is the negative logarithm to the base 10 of the concentration of

H+ ions (the same definition is right for the pOH). So the pH grows and shrinks in a

logarithm scale, while at the opposite the pOH shrinks and grows complementary.

Example

Calculate the pH of 0.06 mol/L HCl.

pH = −log0.06 = 1.22

You can see some examples how to calculate ph and about oxidation numbers and vectors

watching the presentation prepared by the Belgian students here: LINK

http://prezi.com/xxfgqfrkirth/maths-in-chemistry/

Links:

www.en.wikipedia.org

www.bbc.co.uk/schools/gcsebitesize/science/

www.towson.edu

Maths and Chemistry chapter

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