06.1 - chemical formulas and composition stoichiometry · chemical formulas and ... the mass...

Chemical formulas and composition stoichiometry

Stoichiometry

The word “stoichiometry” is derived from the Greek stoicheion, which means “element,” and metron, which means “measure.”

Stoichiometry describes the quantitative relationships among elements in

compounds (composition stoichiometry) and among substances as they

undergo chemical changes (reaction stoichiometry).

Main division of chemical compounds: organic vs inorganic

Many of the molecules found in nature are organic compounds.

• Organic compounds contain C-C or C-H bonds or both, often in combination with nitrogen, oxygen, sulfur, and other elements.

• All of the other compounds are inorganic compounds.

Name Formula Name Formula Name Formula

water H2O sulfur dioxide SO2 butane C4H10

hydrogen peroxide H2O2 sulfur trioxide SO3 pentane C5H12

hydrogen chloride* HCl carbon monoxide CO benzene C6H6

sulfuric acid H2SO4 carbon dioxide CO2 methanol (methyl alcohol) CH3OH

nitric acid HNO3 methane CH4 ethanol (ethyl alcohol) CH3CH2OH

acetic acid CH3COOH ethane C2H6 acetone CH3COCH3

ammonia NH3 propane C3H8 diethyl ether (ether) CH3CH2OCH2CH3

Chemical formulasThe chemical formula gives the number of atoms of each type in the molecule. But this formula does not express the order in which the atoms in the molecules are bonded together.

C C C HH

H

H H

H

H

H

The structural formula shows the order in which atoms are connected. The lines connecting atomic symbols represent chemical bonds between atoms

The structural formula of propane shows that the three C atoms are linked in a chain, with three H atoms bonded to each of the end C atoms and two H atoms bonded to the center C.

For instance, the structural formula of propane is C3H8 and shows that the molecule is composed of 3 carbons and 8 hydrogens.

Chemists sometimes write out the formula in a way to better convey the connectivity information, e.g., CH3CH2CH3, which is a longer representation of the propane chemical formula.

Chemical formulas

Each molecule of acetic acid, CH3COOH, contains 2 carbon atoms, 4 hydrogen atoms, and 2 oxygens.

Writing it as CH3COOH (instead of C2H4O2) includes useful bonding and structural information.

C C

H

H

H

O

OH

Law of Definite Proportions

A compound can be decomposed by chemical means into simpler substances, always in the same ratio by mass.

E.g.

Water originates from the reaction between two H and one O. This leads to the ratio 1:8 which is the ratio of the masses of H and O in the molecule of H2O.

AWH = 1,0079 amu = 1AWO = 15,9994 amu = 16

One oxygen is sixteen times the mass of a H. Hence, in the molecule of water

Water is 11.1% hydrogen and 88.9% oxygen by mass.

H2

2 amu

O16 amu

H2O

FW = 181 8:

Law of Definite Proportions

Carbon dioxide (CO2) is 27.3% carbon and 72.7% oxygen by massCalcium oxide (CaO) is 71.5% calcium and 28.5% oxygen by massCalcium carbonate (CaCO3) is 40.1% calcium, 12.0% carbon, 47.9% oxygen by mass

Observations such as these on many pure compounds (as CaCO3) led Joseph-Louis Proust in1799 to the statement of the Law of Definite Proportions (also known as the Law of Constant Composition):

Different samples of a compound always contain the same elements in the same proportion by mass; this corresponds to atoms of these elements combined in fixed numerical ratios.

Percent composition and chemical formulas

If the chemical formula is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition).

E.g. One CO2, contains one C atom and two O atoms. Percentage is the part divided by the whole times 100%, so we can represent the percent composition of carbon dioxide as follows:

60 C H A P T E R 2 • C H E M I C A L F O R M U L A S A N D C O M P O S I T I O N S T O I C H I O M E T R Y

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2-7 Percent Composition and Formulas of CompoundsIf the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition). For example, one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is the part divided by the whole times 100% (or simply parts per hundred), so we can rep-resent the percent composition of carbon dioxide as follows:

%C 5mass of C

mass of CO23 100% 5

AW of CMW of CO2

3 100% 512.0 amu44.0 amu

3 100% 5 27.3% C

%O 5mass of O

mass of CO23 100% 5

2 3 AW of OMW of CO2

3 100% 52 (16.0 amu)

44.0 amu3 100% 5 72.7% O

One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms (32.0 g). We could therefore have used these masses in the preceding calculation. These numbers are the same as the ones usedionly the units are different. In Example 2-12 we will base our calculation on one mole rather than one molecule.

▶ AW 5 atomic weight (mass)MW 5 molecular weight (mass)

As a check, we see that the percentages add to 100%.

Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. All samples of pure HNO3 have this composition, according to the Law of Definite Proportions.

Calculate the percent composition by mass of HNO3.

PlanWe first calculate the mass of one mole as in Example 2-8. Then we express the mass of each element as a percent of the total.

SolutionThe molar mass of HNO3 is calculated first.

Number of Mol of Atoms 3 Mass of One Mol of Atoms 5 Mass Due to Element

1 3 H 5 1 3 1.0 g 5 1.0 g of H 1 3 N 5 1 3 14.0 g 5 14.0 g of N 3 3 O 5 3 3 16.0 g 5 48.0 g of O

Mass of 1 mol of HNO3 5 63.0 gNow, its percent composition is

% H 5mass of H

mass of HNO33 100% 5

1.0 g63.0 g

3 100% 5 1.6% H

% N 5mass of N


14.0 g63.0 g

3 100% 5 22.2% N

% O 5mass of O


48.0 g63.0 g

3 100% 5 76.2% O

Total 5 100.0%

You should now work Exercise 62.

EXAMPLE 2-12 Percent Composition

▶ When chemists use the % notation, they mean percent by mass unless they specify otherwise.

10663_02_ch02_p043-080.indd 60 12/7/12 3:02 PM

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2-7 Percent Composition and Formulas of CompoundsIf the formula of a compound is known, its chemical composition can be expressed as the mass percent of each element in the compound (percent composition). For example, one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is the part divided by the whole times 100% (or simply parts per hundred), so we can rep-resent the percent composition of carbon dioxide as follows:

%C 5mass of C

mass of CO23 100% 5

AW of CMW of CO2

3 100% 512.0 amu44.0 amu

3 100% 5 27.3% C

%O 5mass of O

mass of CO23 100% 5

2 3 AW of OMW of CO2

3 100% 52 (16.0 amu)

44.0 amu3 100% 5 72.7% O

One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms (32.0 g). We could therefore have used these masses in the preceding calculation. These numbers are the same as the ones usedionly the units are different. In Example 2-12 we will base our calculation on one mole rather than one molecule.

▶ AW 5 atomic weight (mass)MW 5 molecular weight (mass)

As a check, we see that the percentages add to 100%.

Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. All samples of pure HNO3 have this composition, according to the Law of Definite Proportions.

Calculate the percent composition by mass of HNO3.

PlanWe first calculate the mass of one mole as in Example 2-8. Then we express the mass of each element as a percent of the total.

SolutionThe molar mass of HNO3 is calculated first.

Number of Mol of Atoms 3 Mass of One Mol of Atoms 5 Mass Due to Element

1 3 H 5 1 3 1.0 g 5 1.0 g of H 1 3 N 5 1 3 14.0 g 5 14.0 g of N 3 3 O 5 3 3 16.0 g 5 48.0 g of O

Mass of 1 mol of HNO3 5 63.0 gNow, its percent composition is

% H 5mass of H


1.0 g63.0 g

3 100% 5 1.6% H

% N 5mass of N


14.0 g63.0 g

3 100% 5 22.2% N

% O 5mass of O


48.0 g63.0 g

3 100% 5 76.2% O

Total 5 100.0%


EXAMPLE 2-12 Percent Composition

▶ When chemists use the % notation, they mean percent by mass unless they specify otherwise.

10663_02_ch02_p043-080.indd 60 12/7/12 3:02 PM


One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms (32.0 g).


Sample exercise• Calculate the percent composition by mass of HNO3We first calculate the mass of one mole (molar mass). Then we express the mass of each element as a percent of the total.The molar mass of HNO3 is:

Number of Mol of Atoms x Mass of One Mol of Atoms = Mass Due to Element

Now, its percent composition is:

•% of H = (1.0 g /63,0 g) 100% = 1,6 %•% of N = (14.0 g /63,0 g) 100% = 22,2 %•% of O = (48,0 g /63,0 g) 100% = 76,2 %

1x H =1 mol1x N =1 mol3x O = 3 mol

x 1,0 gx 14,0 gx 16,0 g

=1,0 g of H= 14,0 g of N= 48,0 g of O

Mass of 1 mol of HNO

RememberThe amount of substance that contains the mass in grams numerically equal to its formula weight in amu contains one mole of the substance. This is the molar mass and is numerically equal to the formula weight and has the units grams/mole (g/mol).


Practice exercise• Calculate the percent composition by mass of each of the following

compoundsDopamine: C8H11NO2

Vitamin E: C29H50O2

Vanillin: C8H8O3

Derivation of Formulas from CompositionThe molecular formula indicates the numbers of atoms in a molecule.

The empirical formula is the smallest whole-number ratio of atoms present.

E.g. The empirical and molecular formulas for water are both H2O; for hydrogen peroxide, the empirical formula is HO, and the molecular formula is H2O2

C2H4O2

C6H12O6

H3PO4

H2OH2O2

CH2OCH2OH3PO4

H2OHO

Molecular formula

Simplest formula

Derivation of Formulas from Composition

One of the first steps in characterizing a new compound is the determination of its percent composition.

A qualitative analysis is performed to determine which elements are present in the compound.

A quantitative analysis is performed to determine the amount of each element.

Once the percent composition of a compound is known, the empirical formula can be determined

Sample exercise (Empirical Formula)• Compounds containing S and O are serious air pollutants and they represent the major

cause of acid rain. A sample of this compound contains 50.1% sulfur and 49.9% oxygen by mass. What is the empirical formula of the compound?

1.Let’s consider 100.0 g of compound, which contains 50.1 g of S and 49.9 g of O. We calculate the number of moles of atoms of each.

2.We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the empirical formula


mol of S =50,1 g

32,1 g mol-1= 1,56 mol of S mol of O =

49,9 g

16 g mol-1= 3,12 mol of O

1,56

1,56= 1,00 S

3.12

1,56= 2,00 O

SO2



▶ A simple and useful way to obtain whole-number ratios among several numbers follows: (a) Divide each number by the smallest number, and then, (b) if necessary, multiply all the resulting numbers by the smallest whole number that will eliminate fractions

▶ The “Relative Mass of Element” column is proportional to the mass of each element in grams. With this interpretation, the next column could be headed “Relative Number of Moles of Atoms.” Then the last column would represent the smallest whole-number ratios of moles of atoms. But because a mole is always the same number of items (atoms), that ratio is the same as the smallest whole-number ratio of atoms.

A 20.882-g sample of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g of O. What is its simplest formula?

PlanWe reason as in Example 2-13, calculating the number of moles of each element and the ratio among them. Here we use the tabu-lar format that was introduced earlier.

Solution Relative Relative Number Convert Fractions Smallest Whole- Mass of of Atoms Divide by to Whole Numbers Number Ratio Element Element (divide mass by AW) Smallest Number (multiply by integer) of Atoms

Na 6.072 6.07223.0

5 0.264 0.2640.264

5 1.00 1.00 3 2 5 2 Na

S 8.474 8.47432.1

5 0.264 0.2640.264

5 1.00 1.00 3 2 5 2 S Na2S2O3

O 6.336 6.33616.0

5 0.396 0.3960.264

5 1.50 1.50 3 2 5 3 O

The ratio of atoms in the simplest formula must be a whole-number ratio (by definition). To convert the ratio 1:1:1.5 to a whole-number ratio, each number in the ratio was multiplied by 2, which gave the simplest formula Na2S2O3 .


EXAMPLE 2-14 Simplest Formula

Step 2: Now we know that 100.0 g of the compound contains 1.56 mol of S atoms and 3.12 mol of O atoms. We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the simplest formula.

1.561.56

5 1.00 S

3.121.56

5 2.00 O

SO2


The solution for Example 2-13 can be set up in tabular form.

Relative Relative Number Smallest Whole- Mass of of Atoms Divide by Number Ratio Element Element (divide mass by AW) Smallest Number of Atoms

S 50.1 50.132.1

5 1.56 1.561.56

5 1.00 S

O 49.9 49.916.0

5 3.12 3.121.56

5 2.00 O SO2

This tabular format provides a convenient way to solve simplest-formula problems, as the next example illustrates.

10663_02_ch02_p043-080.indd 62 12/7/12 3:02 PM


m

mol x Fw

Sample exercise (Empirical Formula)• 20.882 g of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g

of O. What is its empirical formula?




▶ A simple and useful way to obtain whole-number ratios among several numbers follows: (a) Divide each number by the smallest number, and then, (b) if necessary, multiply all the resulting numbers by the smallest whole number that will eliminate fractions

▶ The “Relative Mass of Element” column is proportional to the mass of each element in grams. With this interpretation, the next column could be headed “Relative Number of Moles of Atoms.” Then the last column would represent the smallest whole-number ratios of moles of atoms. But because a mole is always the same number of items (atoms), that ratio is the same as the smallest whole-number ratio of atoms.

A 20.882-g sample of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g of O. What is its simplest formula?

PlanWe reason as in Example 2-13, calculating the number of moles of each element and the ratio among them. Here we use the tabu-lar format that was introduced earlier.

Solution Relative Relative Number Convert Fractions Smallest Whole- Mass of of Atoms Divide by to Whole Numbers Number Ratio Element Element (divide mass by AW) Smallest Number (multiply by integer) of Atoms

Na 6.072 6.07223.0

5 0.264 0.2640.264

5 1.00 1.00 3 2 5 2 Na

S 8.474 8.47432.1

5 0.264 0.2640.264

5 1.00 1.00 3 2 5 2 S Na2S2O3

O 6.336 6.33616.0

5 0.396 0.3960.264

5 1.50 1.50 3 2 5 3 O

The ratio of atoms in the simplest formula must be a whole-number ratio (by definition). To convert the ratio 1:1:1.5 to a whole-number ratio, each number in the ratio was multiplied by 2, which gave the simplest formula Na2S2O3 .


EXAMPLE 2-14 Simplest Formula

Step 2: Now we know that 100.0 g of the compound contains 1.56 mol of S atoms and 3.12 mol of O atoms. We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the simplest formula.

1.561.56

5 1.00 S

3.121.56

5 2.00 O

SO2


The solution for Example 2-13 can be set up in tabular form.

Relative Relative Number Smallest Whole- Mass of of Atoms Divide by Number Ratio Element Element (divide mass by AW) Smallest Number of Atoms

S 50.1 50.132.1

5 1.56 1.561.56

5 1.00 S

O 49.9 49.916.0

5 3.12 3.121.56

5 2.00 O SO2

This tabular format provides a convenient way to solve simplest-formula problems, as the next example illustrates.

10663_02_ch02_p043-080.indd 62 12/7/12 3:02 PM


6,072 g

23,0 g mol-1Na

S

O

6,072 g

8,474 g

6,336 g

8,474 g

32,0 g mol-1

6,336 g

16,0 g mol-1

= 0,264 mol

= 0,264 mol

= 0,396 mol

0,264

0,264= 1,00

0,264

0,264= 1,00

0,396

0,264= 1,50

1,00x2 = 2,00

1,00x2 = 2,00

1,50x2 = 3,00

Empirical formula = Na2S2O3

N.B. The ratio of atoms in the empirical formula must be a whole-number ratio (by definition). To convert the ratio 1:1:1.5 to a whole- number ratio, each number in the ratio was multiplied by 2, which gave the empirical formula Na2S2O3

m

mol x Fw


Practice exercise• The hormone norepinephrine is released in the human body during stress

and increases the body’s metabolic rate. Like many biochemical compounds, norepinephrine is composed of carbon, hydrogen, oxygen, and nitrogen. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. What is the empirical formula of norepinephrine?


Remember: Hydrocarbons are organic compounds composed entirely of hydrogen and carbon.

Sample exercise (Percent composition)• A 0.1647 g sample of hydrocarbon is burned in a C-H combustion train to produce 0.4931 g of CO2

and 0.2691 g of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.

1. With a proportion, we use the observed masses to determine the masses of C and H in the original sample. There is one mole of carbons, (aw=12.01), in each mole of CO2, 44.01 g; there are two moles of hydrogen atoms, (aw=2.016), in each mole of H2O, 18.02 g:

2. Then we calculate the percentages by mass of each element:

12.01 g/mol C

44.01 g/mol CO2

g of C = 0.4931 g CO2 x = 0.1346 g C

2.016 g/mol H

18.02 g/mol H2Og of H = 0.2691 g H2O x = 0.0301 g H

0.1346 g C

0.1647 g sample% C = x 100% = 81.72% C

0.0301 g C

0.1647 g sample% H = x 100% = 18.28%H

mass of element

mass of the compound

AW

FW=

Molecular Formulas from Empirical Formulas To determine the molecular formula of a substance, both its empirical formula and its formula weight must be known.

The molecular formula is a multiple of the empirical formula.

E.g. Butane, C4H10. The empirical formula for butane is C2H5, but the molecular formula contains twice as many atoms; that is, 2x(C2H5)=C4H10.

Benzene, C6H6. The empirical formula for benzene is CH. the molecular formula has six times as many atoms; 6x(CH)=C6H6.

The molecular formula for a compound is either the same as, or an integer multiple of, the (empirical) formula.

molecular weight

empirical formula weighn =so

A practical example: Combustion Analysis One technique for determining empirical formulas in the laboratory is combustion analysis, commonly used for organic compounds.

When a compound containing carbon and hydrogen is combusted, the carbon is converted to CO2 and the hydrogen is converted to H2O. The amounts of CO2 and H2O produced are determined by measuring the mass increase in the CO2 and H2O absorbers. From the masses of CO2 and H2O we can calculate the number of moles of C and H in the original sample and thereby the empirical formula.

SECTION 3.5 Empirical Formulas from Analyses 95

Combustion AnalysisOne technique for determining empirical formulas in the laboratory iscombustion analysis, commonly used for compounds containing princi-pally carbon and hydrogen.

When a compound containing carbon and hydrogen is completelycombusted in an apparatus such as that shown in ! FIGURE 3.14, thecarbon is converted to CO2 and the hydrogen is converted to H2O.•(Section 3.2) The amounts of CO2 and H2O produced are deter-mined by measuring the mass increase in the CO2 and H2O absorbers.From the masses of CO2 and H2O we can calculate the number of molesof C and H in the original sample and thereby the empirical formula. Ifa third element is present in the compound, its mass can be determinedby subtracting the measured masses of C and H from the original sam-ple mass.

O2Sample

FurnaceH2O absorber CO2 absorber

Sample combusted,producing CO2 and H2O

Mass gained by each absorber corresponds to mass ofCO2 or H2O produced

H2O and CO2 are trappedin separate absorbers

" FIGURE 3.14 Apparatus forcombustion analysis.

To calculate the mass of C from the measured mass ofCO2, we first use the molar mass of CO2, 44.0 g>mol, toconvert grams of CO2 to moles of CO2. Because eachCO2 molecule has only one C atom, there is 1 mol of Catoms per mole of CO2 molecules. This fact allows usto convert moles of CO2 to moles of C. Finally, we usethe molar mass of C, 12.0 g, to convert moles of C tograms of C: Grams C = (0.561 g CO2)¢ 1 mol CO2

44.0 g CO2≤ ¢ 1 mol C

1 mol CO2≤ ¢ 12.0 g C

1 mol C≤ = 0.153 g C

The calculation for determining H mass from H2O massis similar, although we must remember that there are 2mol of H atoms per 1 mol of H2O molecules:

Grams H = (0.306 g H2O)¢ 1 mol H2O

18.0 g H2O≤ ¢ 2 mol H

1 mol H2O≤ ¢ 1.01 g H

1 mol H≤ = 0.0343 g H

The mass of the sample, 0.255 g, is the sum of themasses of C, H, and O. Thus, the O mass is = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O

Mass of O = mass of sample - (mass of C + mass of H)

The number of moles of C, H, and O in the sample istherefore

Moles O = (0.068 g O) ¢ 1 mol O16.0 g O

≤ = 0.0043 mol O

Moles H = (0.0343 g H)¢ 1 mol H1.01 g H

≤ = 0.0340 mol H

Moles C = (0.153 g C)¢ 1 mol C12.0 g C

≤ = 0.0128 mol C

To find the empirical formula, we must compare therelative number of moles of each element in the sam-ple. We determine relative number of moles by divid-ing each of our calculated number of moles by thesmallest number:

C:0.01280.0043

= 3.0 H: 0.03400.0043

= 7.9 O: 0.00430.0043

= 1.0

The first two numbers are very close to the wholenumbers 3 and 8, giving the empirical formula C3H8O.

SAMPLE EXERCISE 3.15 Determining an Empirical Formula by Combustion Analysis

Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropylalcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol.

SOLUTIONAnalyze We are told that isopropyl alcohol contains C, H, and Oatoms and given the quantities of CO2 and H2O produced when agiven quantity of the alcohol is combusted. We must determine theempirical formula for isopropyl alcohol, a task that requires us to cal-culate the number of moles of C, H, and O in the sample.

Plan We can use the mole concept to calculate grams of C in the CO2and grams of H in the H2O. These masses are the masses of C and Hin the alcohol before combustion. The mass of O in the compoundequals the mass of the original sample minus the sum of the C and Hmasses. Once we have the C, H, and O masses, we can proceed as inSample Exercise 3.13.

Solve

︎Apparatus for combustion analysis

If a third element is present in the compound, its mass can be determined by subtracting the measured masses of C and H from the original sample mass.

Law of Multiple ProportionsTwo (or more) elements may form more than one compound (e.g. CO and CO2, NO and NO2, H2O and H2O2). The Law of Multiple Proportions states that

“When two elements, A and B, combine and form more than one compound, the ratio of the masses of element B, in each of the compounds, can be

expressed by small whole numbers”

E.g.: SO2 and SO3 provide an example. In SO2, two moles of oxygen combine with one mole of sulfur atomsIn SO3, three moles of oxygen combine with one mole of sulfur atoms.

Thus the ratio of oxygen atoms in the two compounds compared to a given number of sulfur atoms is 2:3.

Many similar examples, such as CO and CO2 (1:2 oxygen ratio) and H2O and H2O2 (1:2 oxygen ratio), are known.

H2

2 amu

O16 amu

H2O

FW = 181 8:

Sample exercise (Empirical Formula)• What is the ratio of the numbers of oxygen atoms that are combined with a given number

of nitrogen atoms in the compounds N2O3 and NO?

1.To compare the number of oxygen atoms, we must have equal numbers of nitrogen atoms.Because NO has half as many nitrogen atoms in its formula relative to N2O3, we must multiply it by a factor of 2 to compare the two elements on the basis of an equal number of nitrogen atoms. Once we show the number of atoms of each element present, we can cancel out the equal amounts of nitrogen atoms, leaving the ratio of oxygen atoms.

The oxygen ratio in the two compounds N2O3 and NO is 3:2

N2O3

2 (NO)=

Law of Multiple Proportions

2N 3O

2N 2O=

3

2Oxygen

ratio=

3O

2O=

Practice exercise• Show that the compounds water, H2O, and hydrogen peroxide, H2O2, obey

the Law of Multiple Proportions.

• Nitric oxide, NO, is produced in internal combustion engines. When NO comes in contact with air, it is quickly converted into nitrogen dioxide, NO2, a very poisonous, corrosive gas. 1. What mass of O is combined with 3.00 g of N in (a) NO and (b) NO2? 2. Show that NO and NO2 obey the Law of Multiple Proportions.

Law of Multiple Proportions

Sample exercise (Empirical Formula)• Once we master the mole concept and the meaning of chemical formulas, we can use them

in many ways. For example, what mass of chromium (Cr) is contained in 35.8 g of (NH4)2Cr2O7?

1.The formula tells us that each mole of (NH4)2Cr2O7 contains two moles of Cr atoms, so we first find the number of moles of (NH4)2Cr2O7. We know that the molar mass is 252.0 g/mol.

2.Each mol of (NH4)2Cr2O7 contains 2 moles of Cr, so we convert the number of moles of (NH4)2Cr2O7 into the number of moles of Cr atoms it contains, using the proportion

3.We then use the atomic weight of Cr to convert the number of moles of chromium atoms to the mass of chromium.

35.8 g252.0 g/mol

==

Other Interpretations of Chemical Formulas

moles of (NH4)2Cr2O7

0.142 mols

moles of Cr

= 0.142 mol (NH4)2Cr2O72 mol Cr

1 mol (NH4)2Cr2O7=x 0.284 mols

g of Cr = 0.284 mol Cr 51.99 g/mol Crx 14.76 g Cr=

m

mol x Fw

Practice exercise• Mercury occurs as a sulfide ore called cinnabar, HgS.

How many grams of mercury are contained in 578 g of pure HgS?

Other Interpretations of Chemical Formulas

06.1 - chemical formulas and composition stoichiometry · chemical formulas and ... the mass...

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