Chapter 3 Calculations with Chemical Formulas and Equations.

Chapter 3Calculations with

Chemical Formulas and

Equations

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Contents and Concepts

Mass and Moles of Substances

Here we will establish a critical relationship between the mass of a chemical substance and the quantity of that substance (in moles).

1. Molecular Mass and Formula Mass

2. The Mole Concept


Determining Chemical Formulas

Explore how the percentage composition and

mass percentage of the elements in a chemical

substance can be used to determine the

chemical formula.

3. Mass Percentages from the Formula

4. Elemental Analysis: Percentages of C, H, and O

5. Determining Formulas


Stoichiometry: Quantitative Relations in

Chemical Reactions

Develop a molar interpretation of chemical

equations, which then allows for calculation

of the quantities of reactants and products.

6. Molar Interpretation of a Chemical Equation

7. Amounts of Substances in a Chemical Equation

8. Limiting Reactant: Theoretical and Percentage Yield


Learning Objectives

Mass and Moles of Substances1. Molecular Mass and Formula Mass

a. Define the terms molecular and formula mass of a substance.

b. Calculate the formula mass from a formula.c. Calculate the formula mass from molecular

models.


2. The Mole Concept

a. Define the quantity called the mole.

b. Learn Avogadro’s number.

c. Understand how the molar mass is related to the formula mass of a substance.

d. Calculate the mass of atoms and molecules.

e. Perform calculations using the mole.

f. Convert from moles of substance to grams of substance.

g. Convert from grams of substance to moles of substance.

h. Calculate the number of molecules in a given mass of a substance.


Determining Chemical Formulas

3. Mass Percentages from the Formula

a. Define mass percentage.

b. Calculate the percentage composition of the elements in a compound.

c. Calculate the mass of an element in a given mass of compound.


4. Elemental Analysis: Percentages of C, H, and O

a. Describe how C, H, and O combustion analysis is performed.

b. Calculate the percentage of C, H, and O from combustion data.


5. Determining Formulas

a. Define empirical formula.

b. Determine the empirical formula of a binary compound from the masses of its elements.

c. Determine the empirical formula from the percentage composition.

d. Understand the relationship between the molecular mass of a substance and its empirical formula mass.

e. Determine the molecular formula from the percentage composition and molecular mass.


Stoichiometry: Quantitative Relations in

Chemical Reactions

6. Molar Interpretation of a Chemical Equation

a. Relate the coefficients in a balanced chemical equation to the number of molecules or moles (molar interpretation).


7. Amounts of Substances in a Chemical Equation

a. Use the coefficients in a balanced chemical equation to perform calculations.

b. Relate the quantities of reactant to the quantity of product.

c. Relate the quantities of two reactants or two products.


8. Limiting Reactant: Theoretical and Percentage Yield

a. Understand how a limiting reactant or limiting reagent determines how many moles of product are formed during a chemical reaction and how much excess reactant remains.

b. Calculate with a limiting reactant involving moles.

c. Calculate with a limiting reactant involving masses.

d. Define and calculate the theoretical yield of chemical reactions.

e. Determine the percentage yield of a chemical reaction.


Molecular MassThe sum of the atomic masses of all the atoms in a molecule of the substance.

Formula MassThe sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not.

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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

2 decimal places74.10 amu


2 significant figures

31.06 amu

Total 74.096

2 O 2(16.00) = 32.00 amu

Ca(OH)2

1 Ca 1(40.08) = 40.08 amu

2 H 2(1.008) = 2.016 amu

CH3NH2

1 C 1(12.01) = 12.01 amu

5 H 5(1.008) = 5.040 amu1 N 1(14.01) = 14.01 amu

Total 31.060

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What is the mass in grams of the nitric acid molecule, HNO3?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.0081 N 1(14.01) = 14.013 O 3(16.00) = 48.00

63.018 (2 decimal places)63.02 g/mol


Mole, molThe 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 = 6.022 × 1023 (to four significant figures).


23

63.02 g 1mol

mol 6.022 10 molecules

221.046843854 10 g

3

22

The mass of one HNO molecule is

1.046 10 g (4 significant figures).

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.


Molar MassThe mass of one mole of substance.

For example:

Carbon-12 has a molar mass of 12 g or 12 g/mol

?


A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.0081 N 1(14.01) = 14.013 O 3(16.00) = 48.00

63.018

(2 decimal places)63.02 g/mol

Note: We need one digit more in the molar mass than in the measured quantity.


63.02 g0.253 mole

1mole

figures)tsignifican(3

g 15.9

Next, using the molar mass, find the mass of 0.253 mole:

= 15.94406 g

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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) = 40.081 C 1(12.01) = 12.013 O 3(16.00) = 48.00

100.09 2 decimal places100.09 g/mol


1mole23.6 g

100.09 g

12.35787791 10 g

12.36 10 g or 0.236 g

(3 significant figures)

Next, find the number of moles in 23.6 g:

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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) = 72.062 O 2(16.00) = 32.001 N 1(14.01) = 14.0114 H 14(1.008) = 14.112

132.1822 decimal places

132.18 g/mol


1mole2.2 g

132.18 g

21.6643 10 mol

21.7 10 mol or 0.017 mol


Next, find the number of moles in 2.2 g:

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The daily requirement of chromium in the human diet is 1.0 × 10-6 g. How many atoms of chromium does this represent?


236 1mol 6.022 10 atoms

1.0 10 g51.996 g 1mol

First, find the molar mass of Cr:1 Cr 1(51.996) = 51.996

Now, convert 1.0 x 10-6 grams to moles:

=1.158166 1016 atoms

1.2 1016 atoms(2 significant figures)

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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) = 207.21 Cr 1(51.996) = 51.9964 O 4(16.00) = 64.00

323.196 (1 decimal place)

323.2 g/mol


207.2 gPb : 100% 64.11%

323.20 g

51.996 gCr : 100% 16.09%

323.20 g

64.00 gO : 100% 19.80%

323.20 g

Now, convert each to percent composition:

Check:64.11 + 16.09 + 19.80 = 100.00

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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) = 144.1211 O 11(16.00) = 176.0022 H 22(1.008) = 22.176

342.296 (2 decimal places)

342.30 g/mol


figures)tsignifican(3

carbong28.7

144.12 g carbon68.1g sucrose

342.30 g sucrose

Now, find the mass of carbon in 68.1 g sucrose:


Percentage CompositionThe 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.

?


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?


Strategy1. 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.


13

1mg3.154637582 10 g C

10 g

232

2 2

1mol CO 1mol C 12.01g C1156 x 10 g CO

44.01g CO 1mol CO 1mol C

= 315.5 mg C


First, find the mass of C in 1156 mg of CO2:


Now, we can find the percentage composition:

26.5 mg100%

342 mg

315.5 mg100%

342 mg C92.3%

H7.7%

342 mg benzene−315.5 mg C

26.5 mg H(the decimal place is not significant; there are 2 significant figures)

Next, find the mass of H in the benzene sample:


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.


Determining the Empirical Formula

Beginning with percent composition:

1. Assume exactly 100 g so percentages convert directly to grams.

2. Convert grams to moles for each element.

3. Manipulate the resulting mole ratios to obtain whole numbers.


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.

?


Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene?

Hmol7.64Hg1.008

Hmol1Hg7.7

Cmol7.685Cg12.01

Cmol1Cg92.3

17.64

7.64

17.64

7.685

Empirical formula: CH


Molecular Formula A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formula.


To determine the molecular formula:

1. Compute the empirical formula weight.

2. Find the ratio of the molecular weight to the empirical formula weight.

3. Multiply each subscript of the empirical formula by n.

weightformulaempirical

weightmolecularn

?


Benzene has the empirical formula CH. Its molecular weight is 78.1 amu. What is its molecular formula?

613.02

78.1

amu13.02weightformulaEmpirical

Molecular formulaC6H6

?


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?

Omol2.631 Og16.00

Omol1Og42.1

Pmol0.7523Pg30.97

Pmol1Pg23.3

Namol1.501Nag22.99

Namol1Nag34.5


Omol2.631

Pmol0.7523

Namol1.501

Empirical formulaNa4P2O7

1.5012.00

0.7523

0.75231.00

0.7523

2.6313.50

0.7523

2 = 4

2 = 2

2 = 7

?


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?

Ng14.01

Nmol1Ng24.1

Hg1.008

Hmol1Hg13.8

Cg12.01

Cmol1Cg62.1

Hmol1.720

Hmol13.69

Cmol5.171


11.720

1.720

81.720

13.69

31.720

5.171

Empirical formulaC3H8N

Hmol1.720

Hmol13.69

Cmol5.171


The empirical formula is C3H8N.

Find the empirical formula weight:

3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu

Molecular formula: C6H16N2

258.10

116 n


StoichiometryThe calculation of the quantities of reactants and products involved in a chemical reaction.

Interpreting a Chemical EquationThe 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.


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

To find the amount of B (one reactant or product) given the amount of A (another reactant or product):

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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?


Molar massesC3H8: 3(12.01) + 8(1.008) = 44.094 gCO2: 1(12.01) + 2(16.00) = 44.01 g

59.9 g CO2


2

2

83

2

83

8383 COmol1

COg44.01

HCmol1

COmol3

HCg44.094

HCmol1HCg20.0

2COg359.8856987

?


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?


Molar masses:

O2 2(16.00) = 32.00 g

C3H8 3(12.01) + 8(1.008) = 44.094 g

2

2

83

2

83

8383 Omol1

Og32.00

HCmol1

Omol5

HCg44.094

HCmol1HCg20.0

2Og572.5722320

72.6 g O2



Limiting ReactantThe 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.


All amounts produced and reacted are determined by the limiting reactant.

How can we determine the limiting reactant?

1. Use each given amount to calculate the amount of product produced.

2. The limiting reactant will produce the lesser or least amount of product.

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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?


Since ZrCl4 gives the lesser amount of Zr, ZrCl4 is the limiting reactant.

0.20 mol Zr will be produced.

Zrmol0.25Mgmol2

Zrmol1Mgmol0.50

Zrmol0.20ZrClmol1

Zrmol1ZrClmol0.20

44

?


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?


Molar massesNH3 1(14.01) + 3(1.008) = 17.02 gCO2 1(12.01) + 2(16.00) = 44.01 gCH4N2O 1(12.01) + 4(1.008) +

2(14.01) + 1(16.00) = 60.06 g

CO2 is the limiting reactant since it gives the lesser amount of CH4N2O.

13.6 g CH4N2O will be produced.

ONCHg13.6

ONCHmol1

ONCHg60.06

COmol1

ONCHmol1

COg44.01

COmol1COg10.0

24

24

24

2

24

2

22

ONCHg17.6

ONCHmol1

ONCHg60.06

NHmol2

ONCHmol1

NHg17.024

NHmol1NHg10.0

24

24

24

3

24

3

33


10.0 at start−7.73 reacted 2.27 g remains

2.3 g NH3 is left unreacted.

(1 decimal place)

reactedNHg7.73 3

3

3

2

3

2

22 NHmol1

NHg17.02

COmol1

NHmol2

COg44.01

COmol1COg10.0

3NHg17.73460577

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:


Theoretical YieldTheorerical yield is the maximum amount of product that can be obtained by a reaction from given amounts of reactants.

This is a calculated amount.


Actual YieldThe amount of product that is actually obtained.

This is a measured amount.

Percentage Yield

actual yieldpercentage yield 100%

theoretical yield

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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?


Theoretical yield = 13.6 gActual yield = 9.3 g

Percent yield =

= 68% yield


9.3 g 100%

13.6 g

Chapter 3 Calculations with Chemical Formulas and Equations.

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Transcript of Chapter 3 Calculations with Chemical Formulas and Equations.