Using Stoichiometry to Determine Chemical Formulas

Q3U2 :Theoretical Yield, Mass Percents, and Empirical Formulas

Theoretical Yield and Actual Yield Theoretical Yield: The amount of product

formed through a chemical reaction, as predicted by Stoichiometry

Actual Yield: The amount of product formed when a reaction is performed in a lab, usually less than theoretical yield.

Why are they different? The collection techniques, apparatus

used, time, and skill of the chemist can effect the actual yield.

Percent Yield

When actual and theoretical yields are different, you express the efficiency of the reaction using percent yield.

Percent yield is the ratio of actual yield to theoretical yield, expressed as a percent.

Percent yield = (actual yield) X 100

(theoretical yield)

Determining Mass Percents Chemical formulas tell you the

elements involved in a reaction, the number of atoms of each and the number of moles of reactants and products

What is the major element in a reaction, by mass?

To answer this, determine the mass percents of each element

Determining Mass Percents Steps1. Determine the molar mass of the

element in the compound2. Multiply the mass of 1 mole times

the number of moles in the compound

3. Now use that mass to find the mass percent in the sample

Example of Mass Percent CalculationsIn 154 grams of C10H18O (Geraniol), How many grams

does the Carbon atom contribute? 1 molecule of C10H18O contains 10 atoms of Carbon Therefore 1 mole of C10H18O contains 10 moles of

carbon atoms Multiply the mass of 1 mole of carbon by 10 to get the

mass of carbon in one mole C10H18O (10 mol) (12.0gC) = 120g C

(1 mol) Use mass to determine mass percent

mass % of C = ( 120 gC ) X 100 = 77.9% (154 g C10H18O)

Lets find the Mass % of the Other AtomsGiven 154 g of C10H18O

What is mass% of H?18 mol H X 1.01 g/mole = 18 g H

Mass% H = mass H X 100 mass C10H18O

= 18.0 g H X 100 = 11.7% H 154 g C10H18O

What is the Mass% of O?

Mass % of O Is?

What is the Mass% of O? (1 mol O) (16 g O) = 16 g O ( 1 mol)

16 g O X 100 = 10.4% O154g C10H18O

Percent Composition, by mass The previous steps will allow you to determine

the composition of any compound, as long as you know the formula.

Try the Following:1. Hydrogen fuels are rated with respect to their

hydrogen content. Determine the percent hydrogen for the following fuels.

a) Ethane, C2H6

b) Methane, CH4

c) Whale Oil, C32H64O2

Answers to percent composition problems

a) Ethane, C2H6 2 mol C, 6 mole H X 12.0g X 1.0 g 24.0g 6.0 g = 30.0g (6.0 g/30.0 g ) X 100 = 20%

b) Methane, CH4 1 mol C , 4 mol H X 12.0g X 1.0 g 12.0 g 4.0 g (4.0 g/16.0 g ) X 100 = 25%

c) Whale Oil, C32H64O2 32 mol C, 64 mol H, 2 mol O X 12.0g X 1.0 g X 16.0 g

384. 0g 64.0 g 32.0 g

(64.0 g/480.0 g ) X 100 = 13%

Practice Percent Composition Calculate the Percent Composition of

oxygen in the following compoundsa. SO3b. CH3COOHc. Ca(NO3)2d. Ammonium Sulfate, (NH3) SO2

Answer % composition

a. 60.00%b. 53.29%c. 58.50% d. 48.43%

More Practice finding mass % composition Hydrogen peroxide (H2O2), what is

the mass % of Oxygen?

Sodium nitrate (NaNO3), what is the mass % of Oxygen?

Answers

a) H2O2 2 mol O, 2 mole H X 16.0g X 1.0

g 32.0g 2.0 g

= 34.0g(32.0 g/34.0 g ) X 100 = 94.1%

b) NaNO3 1 mol Na, 1 mol N, 3 mole O

X 23.0g X 14.0 g X 16.0 g

23.0 g 14.0g 48.0 g

(48.0 g/85.0 g ) X 100 = 56.4%

Determining Empirical Formulas Fish in some lakes have been found to contain a mercury

compound, possibly a contaminant from the making of paper.

Analysis of this compound gives the following mass percentages: Carbon, 5.57% ; hydrogen, 1.40% ; and mercury, 93.03%.

Using this information, determine the Empirical Formula of the compound.

Empirical Formula: the formula of a compound that has the smallest whole-number ratio of atoms in the compound

Determining Empirical Formulas

Steps1. Fin d the relative numbers of atoms in the

formula unit of the compound2. Use the molar mass to find the number of

moles of each element3. Divide the mole numbers by the smallest

one, to find the whole number ratios

Empirical Formula: the formula of a compound that has the smallest whole-number ratio of atoms in the compound

Example Problem: Empirical Formula Analysis of this compound gives the following mass percentages: Carbon,

5.57% ; hydrogen, 1.40% ; and mercury, 93.03%. Using this information, determine the Empirical Formula of the compound.

Carbon: 5.57g 12.01 g/mol = 0.464 mol ratio = 1

Hydrogen: 1.40 g 1.008 g/mol = 1.39 mol ratio: 1.39/

0.464 = 3

Mercury: 93.03g 200.59 g/mol = 0.464 mol ratio = 1

Empirical formula is CH3Mg, methyl mercury

Practice Finding the Empirical Formula An unknown compound is 18.8% Na,

29.0% Cl and 52.2% O, what is the empirical formula of the unknown?

Answer Use molar mass to find the number of moles of each element.

(18.8 g Na) (1mol Na) = 0.817 mol Na (23.0 g Na) (29.0 g Cl) (1mol Cl) = 0.817 mol Cl (35.5 g Cl) (52.2 g O) (1mol O) = 3.26 mol O (16.0 g O)

Divide the mole numbers by the smallest one. 0.817/ 0.817 = 1 mol Na ratio = 1 0.817/ 0.817 = 1 mol Cl ratio = 1 3.26/ 0.817 = 3.99 mol O ratio = 4

The Empirical formula is NaClO4

Determine the Empirical formula Hydrogen = 5.17%, nitrogen is

35.9% and sodium is 58.9%, what is the empirical formula?

Answer

5.17 g H X 1 mol H/ 1.01 g H = 5.12 mol H

35.9 g N X 1 mol N/ 14.0 g N = 2.56 mol N

58.9 g Na X 1 mol Na/ 23.0 g Na = 2.56 mol Na

Ratio of H:N:Na is 2:1:1

Empirical formula is NaNH2

Calculate the empirical formula for the following

a.       15.8% carbon and 84.2% sulfur

b.      43.6% phosphorus and 56.4% oxygen

c.       28.7% K, 1.5% H, 22.8% P and 47.0% O

 

a. CS2 b. P2O5c. KH2PO4

Calculate the empirical formulaCalculate the empirical formula for the

following compounds from percent composition

a.  0.0130 mol C, 0.0390 mol H, 0.0065 mol O

b.  11.66 g iron, 5.01 g oxygen

c.  40.0 percent C, 6.7 percent H, and 53.3 percent O by mass

Answers: Empirical Formulaa. C2H6Ob. Fe2O3 c. CH2O

From the Empirical Formula to the Chemical Formula For most ionic compounds the two are the

same Covalent compounds can be a very

different case. The ratio between moles is the same, but the

atoms can share electrons differently, so the formula may be different. ▪ Ex. C6H12O6 and C3H6O3 and C2H4O2 glucose lactic acid acetic acidDifferent compounds with different properties, same

molar ratios, different chemical formulas!

From Empirical Formula to Chemical Formula You need the molar mass of the

compound, change to molecular mass (amu)

Divide the molecular mass of the compound by the molecular mass of the empirical formula unit

This will tell you how many of the empirical formula units are in the chemical formula

You try it!

40.0% C, 6.7% H, 53.3% OFind the empirical formula

Is this the same as the chemical formula, if the molar mass is 90.0 g/mol

answer 3.33 mol C = 1 mol C 6.7o mol H = 2.01 mol H 3.33 mol O = 1.o mol OEmpirical formula is CH2O

90.0 g/mol = 90.0 amu molecular massCH2O molecular mass= 12+2 + 16= 30 amu

90.0/30.0 =3 3 empirical formula units C3H6O3, lactic acid

Calculate the chemical formula for the following a.    empirical formula CH , molar mass = 78 g/mol  b.    empirical formula NO2 , molar mass = 92.02 g/mol  c      caffeine, 49.5% C , 5.15% H , 28.9% N , 16.5%

O by mass, molar mass = 195 g.   

answers

a. C6H6

b. N2O4

c. Use mlar mass to find ratios, and empirical formula first, divide molecular and empirical formula masses to determine number of empirical formula units in sample grams C8H10N4O2