1. Chapter 3 Calculations with Chemical Formulas and Equations

2. Molecular Mass Formula Mass
The sum of the atomic masses of all the atoms in a molecule of the substance.
Formula Mass
The sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not.
Copyright © Cengage Learning. All rights reserved.

3. calcium hydroxide, Ca(OH)2 methylamine, CH3NH2
Calculate the formula weight of the following compounds from their formulas. Report your answers to three significant figures.
calcium hydroxide, Ca(OH)2
methylamine, CH3NH2
Copyright © Cengage Learning. All rights reserved.

4. Ca(OH)2 1 Ca 1(40.08) = 40.08 amu 2 O 2(16.00) = 32.00 amu
2 H 2(1.008) = amu
2 decimal places
74.10 amu
Total
CH3NH2
1 C 1(12.01) = amu
1 N 1(14.01) = amu
5 H 5(1.008) = amu
Total
2 significant figures
31.06 amu
Copyright © Cengage Learning. All rights reserved.

5. What is the mass in grams of the nitric acid molecule, HNO3?
First, find the molar mass of HNO3:
1 H 1(1.008) = 1.008
1 N 1(14.01) = 14.01
3 O 3(16.00) = 48.00
63.018
(2 decimal places)
63.02 g/mol
Copyright © Cengage Learning. All rights reserved.

6. NA = 6.022 × 1023 (to four significant figures).
Mole, mol
The quantity of a given amount of substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of carbon-12.
Avogadro’s Number, NA
The number of atoms in exactly 12 g of carbon-12
NA = × 1023 (to four significant figures).
Copyright © Cengage Learning. All rights reserved.

7. Next, convert this mass of one mole to one molecule using Avogadro’s number:
Note:
We need as many significant figures (or one extra) in the constant as in the measured quantity.
Copyright © Cengage Learning. All rights reserved.

8. Molar Mass The mass of one mole of substance. For example:
Carbon-12 has a molar mass of 12 g or 12 g/mol
Copyright © Cengage Learning. All rights reserved.

9. A sample of nitric acid, HNO3, contains 0. 253 mol HNO3
A sample of nitric acid, HNO3, contains mol HNO3. How many grams is this?
Note:
We need one digit more in the molar mass than in the measured quantity.
First, find the molar mass of HNO3:
1 H 1(1.008) = 1.008
1 N 1(14.01) = 14.01
3 O 3(16.00) = 48.00
63.018
(2 decimal places)
63.02 g/mol
Copyright © Cengage Learning. All rights reserved.

10. Next, using the molar mass, find the mass of 0.253 mole:
Copyright © Cengage Learning. All rights reserved.

11. Calcite is a mineral composed of calcium carbonate, CaCO3
Calcite is a mineral composed of calcium carbonate, CaCO3. A sample of calcite composed of pure calcium carbonate weighs 23.6 g. How many moles of calcium carbonate is this?
First, find the molar mass of CaCO3:
1 Ca 1(40.08) = 40.08
1 C 1(12.01) = 12.01
3 O 3(16.00) = 48.00
100.09
2 decimal places
g/mol
Copyright © Cengage Learning. All rights reserved.

12. Next, find the number of moles in 23.6 g:
Copyright © Cengage Learning. All rights reserved.

13. The average daily requirement of the essential amino acid leucine, C6H14O2N, is 2.2 g for an adult. What is the average daily requirement of leucine in moles?
First, find the molar mass of leucine:
6 C 6(12.01) = 72.06
2 O 2(16.00) = 32.00
1 N 1(14.01) = 14.01
14 H 14(1.008) =
2 decimal places
g/mol
Copyright © Cengage Learning. All rights reserved.

14. Next, find the number of moles in 2.2 g:
Copyright © Cengage Learning. All rights reserved.

15. The daily requirement of chromium in the human diet is 1. 0 × 10-6 g
The daily requirement of chromium in the human diet is 1.0 × 10-6 g. How many atoms of chromium does this represent?
Copyright © Cengage Learning. All rights reserved.

16. (2 significant figures)
First, find the molar mass of Cr:
1 Cr 1(51.996) =
Now, convert 1.0 x 10-6 grams to moles:
=  1016 atoms
1.2  1016 atoms
(2 significant figures)
Copyright © Cengage Learning. All rights reserved.

17. The chemical name of table sugar is sucrose, C12H22O11
The chemical name of table sugar is sucrose, C12H22O11. How many grams of carbon are in 68.1 g of sucrose?
First, find the molar mass of C12H22O11:
12 C 12(12.01) =
11 O 11(16.00) =
22 H 22(1.008) =
(2 decimal places)
g/mol
Copyright © Cengage Learning. All rights reserved.

18. Now, find the mass of carbon in 68.1 g sucrose:
Copyright © Cengage Learning. All rights reserved.

19. Percentage Composition
The mass percentage of each element in the compound.
The composition is determined by experiment, often by combustion. When a compound is burned, its component elements form oxides—for example, CO2 and H2O. The CO2 and H2O are captured and weighed to determine the amount of C and H in the original compound.
Copyright © Cengage Learning. All rights reserved.

20. Benzene is a liquid compound composed of carbon and hydrogen; it is used in the preparation of polystyrene plastic. A sample of benzene weighing 342 mg is burned in oxygen and forms 1156 mg of carbon dioxide. What is the percentage composition of benzene?
Copyright © Cengage Learning. All rights reserved.

21. 1. Use the mass of CO2 to find the mass of carbon from the benzene.
Strategy
1. Use the mass of CO2 to find the mass of carbon from the benzene.
2. Use the mass of benzene and the mass of carbon to find the mass of hydrogen.
3. Use these two masses to find the percent composition.
Copyright © Cengage Learning. All rights reserved.

22. First, find the mass of C in 1156 mg of CO2:
= mg C
(4 significant figures)
Copyright © Cengage Learning. All rights reserved.

23. Next, find the mass of H in the benzene sample:
342 mg benzene
−315.5 mg C
26.5 mg H
(the decimal place is not significant; there are 2 significant figures)
Now, we can find the percentage composition:
Copyright © Cengage Learning. All rights reserved.

24. Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow)
Lead(II) chromate, PbCrO4, is used as a paint pigment (chrome yellow). What is the percentage composition of lead(II) chromate?
First, find the molar mass of PbCrO4:
1 Pb 1(207.2) = 207.2
1 Cr 1(51.996) =
4 O 4(16.00) = 64.00
(1 decimal place)
323.2 g/mol
Copyright © Cengage Learning. All rights reserved.

25. Now, convert each to percent composition:
Check: =
Copyright © Cengage Learning. All rights reserved.

26. Empirical Formula (Simplest Formula)
The formula of a substance written with the smallest integer subscripts.
For example:
The empirical formula for N2O4 is NO2.
The empirical formula for H2O2 is HO.
Copyright © Cengage Learning. All rights reserved.

27. Determining the Empirical Formula Beginning with percent composition:
Assume exactly 100 g so percentages convert directly to grams.
Convert grams to moles for each element.
Manipulate the resulting mole ratios to obtain whole numbers.
Copyright © Cengage Learning. All rights reserved.

28. Manipulating the ratios:
Divide each mole amount by the smallest mole amount.
If the result is not a whole number:
Multiply each mole amount by a factor to make whole numbers.
For example:
If the decimal portion is 0.5, multiply by 2.
If the decimal portion is 0.33 or 0.67, multiply by 3.
If the decimal portion is 0.25 or 0.75, multiply by 4.
Copyright © Cengage Learning. All rights reserved.

29. Benzene is composed of 92. 3% carbon and 7. 7% hydrogen
Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene?
Empirical formula: CH
Copyright © Cengage Learning. All rights reserved.

30. Molecular Formula
A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula.
Copyright © Cengage Learning. All rights reserved.

31. To determine the molecular formula:
Compute the empirical formula weight.
Find the ratio of the molecular weight to the empirical formula weight.
Multiply each subscript of the empirical formula by n.
Copyright © Cengage Learning. All rights reserved.

32. Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula?
Molecular formula
C6H6
Copyright © Cengage Learning. All rights reserved.

33. Sodium pyrophosphate is used in detergent preparations
Sodium pyrophosphate is used in detergent preparations. It is composed of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula?
Copyright © Cengage Learning. All rights reserved.

34.  2 = 4  2 = 2  2 = 7 Empirical formula Na4P2O7
Copyright © Cengage Learning. All rights reserved.

35. Hexamethylene is one of the materials used to produce a type of nylon
Hexamethylene is one of the materials used to produce a type of nylon. It is composed of 62.1% C, 13.8% H, and 24.1% N. Its molecular weight is 116 amu. What is its molecular formula?
Copyright © Cengage Learning. All rights reserved.

36. Empirical formula
C3H8N
Copyright © Cengage Learning. All rights reserved.

37. Molecular formula: C6H16N2
The empirical formula is C3H8N.
Find the empirical formula weight:
3(12.01) + 8(1.008) + 1(14.01) = amu
Molecular formula: C6H16N2
Copyright © Cengage Learning. All rights reserved.

38. Interpreting a Chemical Equation
Stoichiometry
The calculation of the quantities of reactants and products involved in a chemical reaction.
Interpreting a Chemical Equation
The coefficients of the balanced chemical equation may be interpreted in terms of either (1) numbers of molecules (or ions or formula units) or (2) numbers of moles, depending on your needs.
Copyright © Cengage Learning. All rights reserved.

39. 1. Convert grams of A to moles of A  Using the molar mass of A
To find the amount of B (one reactant or product) given the amount of A (another reactant or product):
1. Convert grams of A to moles of A
 Using the molar mass of A
2. Convert moles of A to moles of B
 Using the coefficients of the balanced chemical equation
3. Convert moles of B to grams of B
 Using the molar mass of B
This should be Figure 3.11 from p. 107 (with no caption).
The currect figure is from the 9e.
Copyright © Cengage Learning. All rights reserved.

40. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of CO2 are produced when 20.0 g of propane is burned?
Copyright © Cengage Learning. All rights reserved.

41. (3 significant figures)
Molar masses
C3H8: 3(12.01) + 8(1.008) = g
CO2: 1(12.01) + 2(16.00) = g
59.9 g CO2
(3 significant figures)
Copyright © Cengage Learning. All rights reserved.

42. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of O2 are required to burn 20.0 g of propane?
Copyright © Cengage Learning. All rights reserved.

43. (3 significant figures)
Molar masses:
O2 2(16.00) = g
C3H8 3(12.01) + 8(1.008) = g
72.6 g O2
(3 significant figures)
Copyright © Cengage Learning. All rights reserved.

44. Limiting Reactant
The reactant that is entirely consumed when a reaction goes to completion.
Once one reactant has been completely consumed, the reaction stops.
Any problem giving the starting amount for more than one reactant is a limiting reactant problem.
Copyright © Cengage Learning. All rights reserved.

45. How can we determine the limiting reactant?
All amounts produced and reacted are determined by the limiting reactant.
How can we determine the limiting reactant?
Use each given amount to calculate the amount of product produced.
The limiting reactant will produce the lesser or least amount of product.
Copyright © Cengage Learning. All rights reserved.

46. ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s)
Magnesium metal is used to prepare zirconium metal, which is used to make the container for nuclear fuel (the nuclear fuel rods):
ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s)
How many moles of zirconium metal can be produced from a reaction mixture containing 0.20 mol ZrCl4 and 0.50 mol Mg?
Copyright © Cengage Learning. All rights reserved.

47. Since ZrCl4 gives the lesser amount of Zr, ZrCl4 is the limiting reactant.
0.20 mol Zr will be produced.
Copyright © Cengage Learning. All rights reserved.

48. Urea, CH4N2O, is used as a nitrogen fertilizer
Urea, CH4N2O, is used as a nitrogen fertilizer. It is manufactured from ammonia and carbon dioxide at high pressure and high temperature:
2NH3 + CO2(g)  CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3 and 10.0 g CO2 were added to a reaction vessel. What is the maximum quantity (in grams) of urea that can be obtained? How many grams of the excess reactant are left at the end of the reactions?
Copyright © Cengage Learning. All rights reserved.

49. Molar masses NH3 1(14.01) + 3(1.008) = 17.02 g
CO2 1(12.01) + 2(16.00) = g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) + 1(16.00) = g
CO2 is the limiting reactant since it gives the lesser amount of CH4N2O.
13.6 g CH4N2O will be produced.
Copyright © Cengage Learning. All rights reserved.

50. Now subtract the amount reacted from the starting amount:
To find the excess NH3, we need to find how much NH3 reacted. We use the limiting reactant as our starting point.
Now subtract the amount reacted from the starting amount:
10.0 at start
−7.73 reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place)
Copyright © Cengage Learning. All rights reserved.

51. Theoretical Yield
Theorerical yield is the maximum amount of product that can be obtained by a reaction from given amounts of reactants.
This is a calculated amount.
Copyright © Cengage Learning. All rights reserved.

52. Actual Yield Percentage Yield
The amount of product that is actually obtained.
This is a measured amount.
Percentage Yield
Copyright © Cengage Learning. All rights reserved.

53. 2NH3 + CO2(g)  CH4N2O + H2O
When 10.0 g NH3 and 10.0 g CO2 are added to a reaction vessel, the limiting reactant is CO2. The theoretical yield is 13.6 of urea. When this reaction was carried out, 9.3 g of urea was obtained. What is the percent yield?
Copyright © Cengage Learning. All rights reserved.

54. (2 significant figures)
Theoretical yield = 13.6 g
Actual yield = 9.3 g
Percent yield =
= 68% yield
(2 significant figures)
Copyright © Cengage Learning. All rights reserved.