Chemistry Classroom Worksheet - Mrs. Crane's...

Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____

Chemistry BChemical

Calculations

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Chemistry Calculations Learning Goals

Worksheet #1 (Math) I can calculate the percent mass of an element in a compound.

Worksheet #2 (Math) I can calculate the simplest (empirical) formula of a compound.

Worksheet #3 (Math) I can calculate the molecular formulas of a compound.

Worksheet #4 (Math) I can calculate percent mass, simplest (empirical) formulas and molecular formulas when given a mixture of problems.

Chemistry Self Assessment SheetDirections: Record the learning goal and homework from the board each day at the start of class. Self evaluate yourself at the end of class using the scales on the learning goals sheet.

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Date: _______________________ Type of Learning Goal (s) (circle) CONCEPT MATH GRAPH

Learning Goal: I CAN _______________________________________________________________________________________________________________________________________________________________________________How well can I do this at the end of class?_________


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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ Purpose: Calculate the percent composition of the two parts of an Oreo cookie. In this lab, we will treat the Oreo cookie as one compound and the two parts (the cookie and the filling) as the two different elements.

Background Information: Percent composition of a compound tells you the percent (by mass) of each element in the compound.

We find percent composition we compare the mass (grams) of each element in a compound to the entire mass (grams) of the whole compound. Then we multiply by 100 to turn the number into a percent.

Pre-Lab QuestionSuppose you sleep seven hours each night. What percentage of each day do you spend sleeping? (Hint: Divide the number

of hours you spend sleeping by the total number of hours in a day. Then multiply the answer by 100 to get the percentage). Show your work below:

Materials: a balance, a metal scoop, a cookie and your brain.

Procedure: You have to write it! Be very specific and very clear so that someone could repeat the lab the EXACT same way you did it. If you don’t have at least 5 steps, you are not being specific enough.

Calculations: Show ALL of your work to determine the percent of each part of the cookie:

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ Worksheet#1: Calculation of Percent by Mass

An extremely common calculation in chemistry is to find the percent by mass of a component in a compound or a mixture. As you know, a percent is calculated by making a fraction of “what you are talking about” over the “whole thing” and then multiplying the fraction by 100.

Example 1: Calculate the percent (%) oxygen by mass in aluminum oxide Al2O3

Step 1: Molar Mass of Al2O3 Step 2: Calculate Percent Mass2 (Al) = 2(26.98) = 53.96 % O = 48.00 g O x 100 = 47.077 = 47.08 % O3 (O) = 3(16.00) = 48.00 101.96 g Al2O3

101.96 g/mol

Example 2: Calculate the % phosphate ion in calcium phosphate Ca3(PO4)2

Step 1: MM of PO4

1 (P) = 1(30.97) = 30.974 (O) = 4(16.00) = 64.00 Step 3: Calculate Percent Mass

94.97g/mol % PO4 = 94.97 X 2 =189.94 g PO4 x 100 = 61.24% PO4

Step 2: MM of Ca3(PO4)2 310.18 g Ca3(PO4)2

3 (Ca) = 3(40.08) = 120.242 (PO4

-3) = 2(94.97)= 189.94310.18 g/mol

Solve the following problems showing all work just as in the examples above.1. Calculate the percent by mass for the named element in the following compounds: (see example 1)

a. Sodium in sodium chloride NaCl

b. Silver in silver oxide Ag2O

c. Oxygen in hydrogen peroxide H2O2

2. Calculate the percent by mass for the named polyatomic ion in each of the following compounds: (see example 2)

a. Calcium in calcium borate Ca3(BO3)2

b. Nitrogen in aluminum nitrate Al(NO3)3

c. carbonate CO3 in potassium carbonate K2CO3

d. ammonium NH4 in ammonium sulfate(NH4)2SO4

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ Worksheet #2: Calculation of Simplest Formula (Empirical Formula)

A formula is the ratio of atoms of each element present in a compound; since moles of elements contain equal numbers of atoms it follows that formulas are also mole ratios. A simplest (empirical) formula is the smallest whole number ratio of the atoms of each elements present. For example: H2O says there are two atoms of hydrogen for each atom of oxygen in water.

Example Problem: A compound is found to contain 36.11% calcium and 63.88% chlorine by mass. What is the simplest formula of this compound?

Solution: Assume a 100.00 gram sample (that means you have 36.11 grams of calcium and 63.88 grams of chlorine).

# moles Ca = 36.11 g Ca x 1 mole Ca = 0.90094 moles Ca (keep 4-5 decimals) 40.08 g Ca

# moles Cl = 63.887 g Cl x 1 mole Cl = 1.8019 moles Cl (keep 4-5 decimals) 35.45 g Cl

We then write the formula by dividing through by the smallest number.Ca0.90094 Cl1.8019 = CaCl2

0.90094 0.90094

Calculate the simplest formula for the following compounds showing all work as done in the example:

1. 35.94% aluminum; 64.06% sulfur

2. 74.185% sodium; 25.815% oxygen

3. 77.73% iron; 22.27% oxygen

4. 26.6% potassium; 35.4% chromium; and 38.0% oxygen

5. 43.64% phosphorus; 56.36% oxygen

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ Worksheet #3: Calculation of Molecular Formulas

For the vast majority of ionic compounds the simplest formula is the only formula that is possible. However, with covalent compounds, many times the formula for the molecule is a whole number multiple of the simplest formula.

For example:Simplest formula Molecular formula

Acetylene CH C2H2

Benzene CH C6H6

Propene CH2 C3H6

So we need to learn to calculate the molecular formula.

EXAMPLE: Calculate the simplest and molecular formula for a hydrocarbon that is 85.60% carbon and 14.40% hydrogen and has a molecular weight of 70.1 grams per mole.

A. First we calculate the simplest formula and find out how many grams are in a simplest formula unit.

Assume 100.00 gram sample of the compound. This means you have 85.60 g C and 14.40 g H.# moles of C = 85.60 g C x 1 mole C = 7.12739 moles C

12.01 g C# moles of H = 14.40 g H x 1 mole H = 14.2574 moles H

1.01 g H**Do not round off the number of moles at this last step.

B. Divide by the smallest number of moles:

C7.12739 H14.2574 CH2 so we know the simplest formula us CH2

7.12739 7.12739

Add up the mass of one simplest formula, this is how much one simplest formula weighsMolar Mass of CH2

1 (C) = 1(12.01) = 12.012 (H) = 2(1.01) = 2.02

14.03 grams in one simplest formula

C. Now calculate the actual molecular formula, you are trying to find out the number of times the simplest formula divides into the molecular weight.

Molecular formula weight ÷ simplest formula weight = 4.996 = 5

So the molecular formula is (CH2) x 5 = C5H10

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ EXAMPLE PROBLEM #1Calculate the simplest formula and the molecular formula for a hydrocarbon that is 84.09% carbon and 15.91% hydrogen.

A. Calculate the simplest formula (WORKSHEET #2!!)

B. Calculate the mass (in grams) of a simplest formula

C. Given the molecular weight of 114.24 g/mol, calculate the molecular formula.

Calculate the simplest formula and the molecular formula for each of the following compounds.1. A compound made of nitrogen and oxygen is 46.68% nitrogen and 53.32% oxygen by mass.

a. Calculate the simplest formula

b. Calculate the mass (in grams) of a simplest formula

c. Given the molecular weight of 60.03 g/mol, calculate the molecular formula.

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ 2. A hydrocarbon (carbon and hydrogen) is found to be 92.24% carbon and 7.76% hydrogen.

HINT: List carbon BEFORE hydrogena. Calculate the simplest formula



3. A compound is 43.64% phosphorus and 56.36% oxygen. a. Calculate the simplest formula



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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ 4. A hydrocarbon (carbon and hydrogen) is 82.63% carbon and 17.37% hydrogen.




EXTRA CHALLENGE:5. A compound is found to have 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen by mass.




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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ EXTRA CHALLENGE WORKSHEETWorksheet #4: Limiting Reactants

When chemicals are mixed together to undergo a reaction, they are often mixed in stoichiometric quantities, that is, in exactly the correct amounts so that all of the reactants “run out” and are used up at the same time.

A good analogy is baking chocolate chip cookies. Imagine you want to make as many cookies as you can. You have lots of flour, eggs, sugar, and vanilla, but you only have 1 package of chocolate chips. Since the recipe for a dozen cookies calls for 1 package of chocolate chip cookies, you can only make four dozen cookies. In this case, you had an excess of all the other ingredients and what determined the amount of cookies (product) you could make was the amount of chocolate chips. In this example the chocolate chips were the limiting reactant.

Now let’s look at a chemical example. In the following reaction of making ammonia, two moles of ammonia would be produced by reacting one mole of nitrogen and three moles of hydrogen.

N2(g) + 3H2(g) 2NH3(g)

1 mole 3 moles 2 moles

Now consider an example where the reactants are not in the correct stoichiometric quantities:N2(g) + 3H2(g) 2NH3(g)

2 moles 3 moles ? moles

Since you now have twice as much nitrogen, you will use all of the hydrogen and some nitrogen will be left over (1 mole) so hydrogen limits the amount of product produced. We will call hydrogen the limiting reactant because the reaction will stop when you run out of hydrogen.

Consider another example where the reactants are not in the correct stoichiometric quantities:N2(g) + 3H2(g) 2NH3(g)

1 mole 4 moles ? moles

Since you have more hydrogen than is needed, you will use up all of the nitrogen and some hydrogen will be left over (1 mole) so nitrogen limits the amount of product produced and we call it the limiting reactant. The reaction will stop when you run out of nitrogen.

How do you know that a problem is a limiting reactant problem and how do you solve it once you have identified it? The answer to the first question is you know when you are given a limiting reactant problem when you are given a stoichiometry problem and the masses of both reactants are given. In order to solve the problem you must follow the following steps”

1. Determine the knowns (what you know/information given) and unknown (what you want to know) and record them above the equation.

2. Using “what you know” (known masses), determine TWO possible masses produces. Use stoichiometry and factor label.

3. Compare the two masses and see what reactant is limiting. The limiting reactant runs out first, SO it limits how much product is produced, therefore the smaller amount of product is produced from the limiting reactant.

EXAMPLE:The reaction between solid white phosphorus and oxygen produces tetraphosphorus decaoxide. Determine the mass in grams of P4O10 formed if 25.0 grams of phosphorus (P4) and 50.0 grams of oxygen are combined.

Balanced Equation = P4 + 5O2 P4O10

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ 1. Record 25 g above P4 and 50 g above O2 in the previous equation.2. Calculate the number of moles of each reactant:

# moles of P4 = 25.0 g x 1 mole = 0.202 mol P4

123.9 g

# moles of O2 = 50.0 g x 1 mole = 1.56 mol O2

32.00 g

3. Calculate the actual ratio of available moles of O2 and available moles of P4 (AKA: Make this a ration over 1)

1.56 mol O2 = 7.72 moles O2

0.202 mole P4 1 mole P4

4. Determine the mole ratio of the two reactants from the balanced chemical equation:

5 moles O2

1 mole P4

5. Since 5 moles of oxygen is needed to react and you have 7.72 moles, the oxygen is in excess and P4 is the limiting reactant. Use the moles of P4 (0.202 moles) to finish the problem and calculate the number of grams of P4O10 produced.

# grams of P4O10 = 0.202 mol P4 x 1 mole P4O10 x 283.9 g P4O10 = 57.3 grams of P4O10

1 mole P4 1 mole P4O10

Solve the following problems on a separate sheet of paper following all of the above steps.

1. The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflate an automotive air bag. The two products are sodium oxide and iron. If 100.0 grams of sodium reacts with 100.0 grams of Fe2O3, determine the mass of solid iron produced.

6Na + Fe2O3 3Na2O + 2Fe2. Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6)

and oxygen. Calculate the mass of glucose produced if 88.0 grams of carbon dioxide and 64.0 grams of water are available to a plant growing in sunlight.

6CO2 + 6H2O C6H12O6 + 6O2 3. Disulfur dichloride is used to vulcanize rubber, a process invented by Charles Goodyear. The process

makes rubber harder, stronger, and less likely to become soft when hot or brittle when cold. In the production of disulfur dichloride, molten sulfur reacts with chlorine gas to the following equation: S8 + 4Cl2 4S2Cl2 If 200.0 grams of sulfur reacts with 100.0 grams of chlorine, what mass of disulfur dichloride is produced?

4. Iron is obtained commercially in a blast furnace by the reaction of hematite (Fe2O3) with carbon monoxide. The two products are iron and carbon dioxide. How many grams of iron are produced if 75.0 grams of hematite react with 40.0 grams of carbon monoxide?

Fe2O3 + 3CO 2Fe + 3 CO2

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ EXTRA CHALLENGE WORKSHEETWorksheet #5: Introduction to Hydrates

Hydrates are compounds which contain molecules of water within their crystal structures. In these compounds, the positive and negative ions of the compound are surrounded by a definite number of water molecules when the compound is in its solid crystalline state. The compounds are called hydrates and the water is called water of hydration or water of crystallization. When we write the formulas for these compounds we indicate the number of water molecules in the crystal by following the formula with a dot and then the appropriate number of water. Examples:

copper (II) sulfate pentahydrate CuSO4 * 5H2O (s)

magnesium sulfate heptahydrate MgSO4 * 7H2O (s)

calcium chloride dihydrate CaCl2 * 2H2O (s)

In the writing of formulas for hydrated compounds the dot is NOT a multiplication sign – it is simply the method used to indicate the water of hydration. The (s) is called a state symbol and means the compound is a solid. NOTICE also, the (s) comes at the end of the formula – the whole formula including the molecules of water represent one substance.

When we heat hydrated salts we drive off their water of hydration as water vapor and we are left with the dehydrated salt (or it may be referred to as the anhydrous salt). Dehydrated is a term that means “water has been removed”. Anhydrous is a term that means “no water is present”. Dehydrated and anhydrous are not exactly synonymous; both mean no water present but the dehydrated says that water was once there and has been removed, anhydrous says there is no water present but it doesn’t tell anything about the history of the material.

Many times we will do laboratory experiments where we heat a hydrate to drive off its water of hydration. We can represent this process by an equation with state symbols.

Example: copper (II) sulfate pentahydrate is heated until the anhydrous salt is formed.CuSO4 * 5H2O (s) CuSO4 (s) + 5H2O (g)

Write the equations including state symbols for dehydrating the following substances.1. magnesium sulfate heptahydrate 4. sodium carbonate decahydrate2. calcium chloride dihydrate 5. calcium sulfate dihydrate3. barium chloride dihydrate 6. sodium thiosulfate pentahydrate

1. ___________________________________________________________________

2. ___________________________________________________________________

3. ____________________________________________________________________

4. ____________________________________________________________________

5. ____________________________________________________________________14

Prefixes1 = mono 6 = hexa2 = di 7 = hepta3 = tri 8 = octa4 = tetra 9 = nona5 = penta 10 = deca

Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ EXTRA CHALLENGE WORKSHEETWorksheet #6: Percent Mass of Hydrates

In worksheet #5 we learned that many salts form crystals with water trapped in their crystal structure and that we call salts hydrated and the water is called water of hydration of water of hydration. It is frequently valuable to know what percent by mass of a crystal is water and what percent is actually the anhydrous salt. We can calculate these values from the formula.

Example: Calculate the percent of water by mass in copper (II) sulfate pentahydrate CuSO4 * 5H2O (s)Step #1: MM of CuSO4 * 5H2O (s) Step #2: Calculate Percent MassCu = 63.54 % H2O = %H2O x 100S = 32.06 CuSO4 * 5H2O4 O = 64.005H2O = 90.10 = 90.10 g x 100 = 36.083% H2OTotal = 249.70 g 249.70 g

USE THIS EXACT SET UP TO CALCULATE THE % OF WATER BY MASS FOR EACH OF THE SIX HYDRATES ON WORKSHEET #5

1.

2.

3.

4.

5.

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Chemistry B: Chemistry Calculations Name: ____________________________ Hour: _____ Worksheet #4: Chemical Calculations Review SheetThis review covers: percent composition by mass, simplest formulas and molecular formulas.

Solve the problems on another sheet of paper. Show all work for credit.

1. A compound was found to contain 41.33% arsenic by mass and 58.67% chlorine by mass. Calculate the simplest formula for this compound. (A: arsenic trichloride, AsCl3)

2. A compound contains 8.39% iron and 3.61% oxygen. Calculate the simplest formula. (A: iron(III) oxide, Fe2O3)

3. Calculate the percent by mass for each element in aluminum hydroxide Al(OH)3. (A: 34.59 % Al, 61.53 % O and 3.88% H)

4. A hydrocarbon was found to contain 84.09% carbon by mass and 15.91% hydrogen by mass. (C4H9 and C8H18)



c. Given the molecular weight of 114g/mol, calculate the molecular formula.

EXTRA CHALLENGE PROBLEM:

5. A student was given a sample of scandium oxide Sc2O3 to analyze for the percent composition by mass. She found that a 23.00 gram sample of the scandium oxide contained 15.00 grams of scandium. Calculate the percent mass for each element in this compound. (65.22% Sc and 34.8% O)

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Chemistry Classroom Worksheet - Mrs. Crane's...

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