Chapter 3

Molecules, Compounds, and

Chemical Equations

3.7 Formula Mass versus Molar

mass Formula mass

◦ The average mass of a molecule or formula unit in amu

◦ also known as molecular mass or molecular weight (MW)

◦ whole = sum of the parts!

Molar mass

◦ Total mass of a

compound in gram per 1

mol of its molecules or

formula unit

Molar Mass of Compounds

the relative masses of molecules can be calculated from atomic masses

Formula Mass = 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu

since 1 mole of H2O contains 2 moles of H and 1 mole of O

Molar Mass = 1 mole H2O

= 2(1.01 g H) + 16.00 g O = 18.02 g

so the Molar Mass of H2O is 18.02 g/mole

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Why multiplying 2?

Molar Mass of Na2SO4

Calculate the molar mass of Na2SO4.

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Element Number

of Moles

Atomic Mass Total Mass

in each

element

Na

S

O

Total mass in 1 mol Na2SO4

3.8 Mass Percent Composition Percentage of each element in a compound

◦ By mass

Can be determined from

1. the formula of the compound

2. the experimental mass analysis of the compound

The percentages may not always total to 100% due to rounding

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100%whole

partPercentage

Example - Mass Percent as a

Conversion Factor

Calculate the mass percent of Na in NaCl

Benzaldehyde is 79.2% carbon. What mass of benzaldehyde

contains 19.8 g of C?

Chemical Formulas and Elemental

Composition

chemical formulas have inherent in them relationships between

numbers of atoms and molecules

◦ or moles of atoms and molecules

these relationships can be used to convert between amounts of

constituent elements and molecules

◦ like percent composition

Vol A Grams A Moles A Moles B Grams B

Grams A Moles A Moles B Grams B

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Example

Butane (C4H10) is the liquid fuel in lighters.

a. Determine the number of atoms ratio between carbon and 1

molecule of C4H10

b. Determine the number moles ratio between C and 1 mol of

C4H10

c. How many grams of carbon are present within a lighter

containing 7.5 mL of butane? The density of quid butane is

0.601 g/mL

Empirical Formula

simplest, whole-number ratio of the atoms of elements in a

compound

can be determined from elemental analysis

◦ masses of elements formed when decompose or react compound

combustion analysis

◦ percent composition

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Steps in determine the Empirical

formula Step 1: Obtain the mass of each element (in grams)

E.G 100% = 100g therefore mass percent is the same numerical value in grams

Step 2: Determine the numbers moles of each atom present

◦ Use molar mass of each element

Step 3: Divide the smallest moles by numbers moles of each atom to obtain the closet integer as possible.

◦ if result is within 0.1 of whole number, round to whole number

Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible.

E.g 1.5 x 2 = 3.0 atoms

1.33 x 3 = 3. 99 = 4 atoms

Step 5: Write the result (number atoms) from step 4 as a subscript for the appropriate element.

Example

Determine the empirical formula of stannous fluoride, which

contains 75.7% Sn (118.70g/mol) and the rest fluorine (19.00

g/mol)

An unknown sample gives the following mass percent:

17.5% Na, 39.7% Cr and 42.8% O. What is the empirical

formula?

Molecular Formulas The molecular formula is a multiple of the empirical formula

To determine the molecular formula you need to know the

empirical formula and the molar mass of the compound

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Multiple (n) =molecular mass

empirical formula mass

Molecular formula = empirical formula x n

where n = 1, 2, 3, 4

Example

Laboratory analysis of aspirin determined the following mass

percent composition. Find the empirical formula and molecular

formula

C = 60.00%

H = 4.48%

O = 35.53%

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Determining Empirical Formulas:

Elemental AnalysisCombustion Analysis: A compound of unknown composition

(containing a combination of carbon, hydrogen, and possibly

oxygen) is burned with oxygen to produce the volatile combustion

products CO2 and H2O, which are separated and weighed by an

automated instrument called a gas chromatograph.

hydrocarbon + O2(g) xCO2(g) + yH2O(g)

carbon

hydrogen

Combustion Analysis

Unknown formula: CxHyOx (Oxygen can be replaced with other

nonmetal)

gCO2 moles CO2 moles C gC

gH2O moles H2O moles H gH

g O = g sample – (g H + g C)

◦ gO moles O

Follow steps in determine the empirical formula and molecular

formula

Example Combustion of a 0.8233 g sample of a compound containing only

carbon, hydrogen, and oxygen produced the following:

CO2 = 2.445 g

H2O = 0.6003 g

Determine the empirical formula of the compound

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Example

Upon combustion, a compound containing only carbon and

hydrogen produced 1.60g CO2 and 0.819g H2O. Find the empirical

formula

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Chemical Reactions

Reactions involve chemical changes in matter resulting in new

substances

Reactions involve rearrangement and exchange of atoms to produce

new molecules

◦ Elements are not transmuted during a reaction

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Chemical Equations

A chemical equation gives

• the formulas of the reactants on the left of the arrow.

• the formulas of the products on the right of the arrow.

Reactants Product

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C(s)

O2 (g)CO2 (g)

Symbols Used in Equations

Symbols in chemical

equations show

• the states of the

reactants.

• the states of the

products.

• the reaction conditions.

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TABLE

Chemical Equations are Balanced

In a balanced

chemical reaction

• no atoms are lost or gained.

• the number of reacting atoms is equal to the number of product atoms.

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Balancing Chemical Equations

A balanced chemical equation shows that the law of conservation

of mass is adhered to.

In a balanced chemical equation, the numbers and kinds of atoms

on both sides of the reaction arrow are identical.

2NaCl(s)2Na(s) + Cl2(g)

right side:

2 Na

2 Cl

left side:

2 Na

2 Cl

Balancing Chemical Equations

2. Find suitable coefficients—the numbers placed before

formulas to indicate how many formula units of each

substance are required to balance the equation.

2H2O(l)2H2(g) + O2(g)

1. Write the unbalanced equation using the correct chemical

formula for each reactant and product.

H2O(l)H2(g) + O2(g)

3. Reduce the coefficients to their smallest whole-number

values, if necessary, by dividing them all by a common

denominator.

2H2O(l)2H2(g) + O2(g)

Balancing Chemical Equations

4. Check your answer by making sure that the numbers and

kinds of atoms are the same on both sides of the equation.

2H2O(l)2H2(g) + O2(g)

right side:

4 H

2 O

left side:

4 H

2 O

Balancing Chemical Equations

Do not change subscripts when you balance a chemical equation.

You are only allowed to change the coefficients.

H2O(l)H2(g) + O2(g) unbalanced

2H2O(l)2H2(g) + O2(g)

Chemical equation changed!

H2O2(l)H2(g) + O2(g)

Balanced properly

Examples

Balance the coefficients from reactants to products.

A. __N2(g) + __H2(g) __ NH3(g)

A. B. __Co2O3(s) + __ C(s) __Co(s) + __CO2(g)

Write a balanced equation for the reaction between

a. carbon dioxide gas and aqueous potassium hydroxide to form

potassium carbonate and water.

b. The combustion of gaseous ethane (C2H6)

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