Quantitative Composition of Compounds Chapter 7 Hein and Arena Eugene Passer Chemistry Department Bronx Community College John Wiley and Sons, Inc. Version 1.1

Chapter Outline7.1 The Mole 7.2 Molar Mass of Compounds7.3 Percent Composition of Compounds7.4 Empirical Formula versus Molecular Formula7.5 Calculating Empirical Formulas7.6 Calculating the Molecular Formula from the Empirical Formula

The Mole

The mass of a single atom is too small to measure on a balance.mass of hydrogen atom = 1.673 x 10-24 g

This is an

infinitesimal

mass1.673 x 10-24 g

Chemists require a unit for counting which can express large numbers of atoms using simple numbers.Chemists have chosen a unit for counting atoms.That unit is theMOLE

1 mole = 6.022 x 1023 objects

LARGE6.022 x 1023is a verynumber

6.022 x 1023isnumberAvogadros Number

If 10,000 people started to count Avogardros number and counted at the rate of 100 numbers per minute each minute of the day, it would take over 1 trillion years to count the total number.

1 mole of any element contains 6.022 x 1023 particles of that substance.

The atomic mass in grams of any element23 contains 1 mole of atoms.

This is the same number of particles 6.022 x 1023 as there are in exactly 12 grams of

Examples

Species

Quantity

Number of H atomsH1 mole6.022 x 1023

Species

Quantity

Number of H2 moleculesH21 mole6.022 x 1023

Species

Quantity

Number of Na atomsNa1 mole6.022 x 1023

Species

Quantity

Number of Fe atomsFe1 mole6.022 x 1023

Species

Quantity

Number of C6H6 moleculesC6H61 mole6.022 x 1023

1 mol of atoms =6.022 x 1023 atoms6.022 x 1023 molecules6.022 x 1023 ions1 mol of molecules =1 mol of ions =

The molar mass of an element is its atomic mass in grams.

It contains 6.022 x 1023 atoms (Avogadros number) of the element.

ElementAtomic massMolar massNumber of atomsH1.008 amu1.008 g6.022 x 1023Mg24.31 amu24.31 g6.022 x 1023Na22.99 amu22.99 g6.022 x 1023

Problems

How many moles of iron does 25.0 g of iron represent?Atomic mass iron = 55.85Conversion sequence: grams Fe moles FeSet up the calculation using a conversion factor between moles and grams.

Atomic mass iron = 55.85Conversion sequence: grams Fe atoms FeHow many iron atoms are contained in 40.0 grams of iron?Set up the calculation using a conversion factor between atoms and grams.

Molar mass Na = 22.99 gConversion sequence: atoms Na grams NaWhat is the mass of 3.01 x 1023 atoms of sodium (Na)?Set up the calculation using a conversion factor between grams and atoms.

What is the mass of 0.365 moles of tin?Atomic mass tin = 118.7Conversion sequence: moles Sn grams SnSet up the calculation using a conversion factor between grams and atoms.

Conversion sequence: moles O2 molecules O atoms OHow many oxygen atoms are present in 2.00 mol of oxygen molecules?Two conversion factors are needed:

Molar Mass of Compounds

The molar mass of a compound can be determined by adding the molar masses of all of the atoms in its formula.

2 C = 2(12.01 g) = 24.02 g6 H = 6(1.01 g) = 6.06 g1 O = 1(16.00 g) = 16.00 g 46.08 gCalculate the molar mass of C2H6O.

1 Li = 1(6.94 g) = 6.94 g1 Cl = 1(35.45 g) = 35.45 g4 O = 4(16.00 g) = 64.00 g106.39 gCalculate the molar mass of LiClO4.

Calculate the molar mass of (NH4)3PO4 .3 N = 3(14.01 g) = 42.03 g12 H = 12(1.01 g) = 12.12 g1 P = 1(30.97 g) = 30.97 g4 O = 4(16.00 g) = 64.00 g149.12 g

Molar Mass1 MOLE

1 MOLE Ca40.078 g Ca

1 MOLE H2OAvogadros Number of H2O molecules6 x 1023 H2O molecules18.02 g H2O

These relationships are present when hydrogen combines with chlorine.

HClHCl6.022 x 1023 H atoms6.022 x 1023 Cl atoms6.022 x 1023 HCl molecules1 mol H atoms1 mol Cl atoms1 mol HCl molecules1.008 g H35.45 g Cl36.46 g HCl1 molar mass H atoms1 molar mass Cl atoms1 molar mass HCl molecules

In dealing with diatomic elements (H2, O2, N2, F2, Cl2, Br2, and I2), distinguish between one mole of atoms and one mole of molecules.

Calculate the molar mass of 1 mole of H atoms.1 H = 1(1.01 g) = 1.01 gCalculate the molar mass of 1 mole of H2 molecules.2 H = 2(1.01 g) = 2.02 g

Problems

How many moles of benzene, C6H6, are present in 390.0 grams of benzene?Conversion sequence: grams C6H6 moles C6H6 The molar mass of C6H6 is 78.12 g.

How many grams of (NH4)3PO4 are contained in 2.52 moles of (NH4)3PO4?Conversion sequence: moles (NH4)3PO4 grams (NH4)3PO4The molar mass of (NH4)3PO4 is 149.12 g.

56.04 g of N2 contains how many N2 molecules?The molar mass of N2 is 28.02 g.Conversion sequence: g N2 moles N2 molecules N2Use the conversion factors

56.04 g of N2 contains how many N2 atoms?The molar mass of N2 is 28.02 g.Conversion sequence: g N2 moles N2 molecules N2 atoms NUse the conversion factors

Percent Compositionof Compounds

Percent composition of a compound is the mass percent of each element in the compound.

Percent Composition From Formula

If the formula of a compound is known a two-step process is needed to calculate the percent composition.Step 1 Calculate the molar mass of the formula.Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

Step 1 Calculate the molar mass of H2S.2 H = 2 x 1.01g = 2.02 g1 S = 1 x 32.07 g = 32.07 g34.09 gCalculate the percent composition of hydrosulfuric acid H2S.

Calculate the percent composition of hydrosulfuric acid H2S.Step 2 Divide the mass of each element by the molar mass and multiply by 100.

Percent Composition From Experimental Data

Percent composition can be calculated from experimental data without knowing the composition of the compound.Step 1 Calculate the mass of the compound formed.Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

Step 1 Calculate the total mass of the compound1.52 g N3.47 g O4.99 gA compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.= total mass of product

Step 2 Divide the mass of each element by the total mass of the compound formed.A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition.

Empirical Formula versus Molecular Formula

The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound.The empirical formula gives the relative number of atoms of each element present in the compound.

The molecular formula is the true formula of a compound.The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

Examples

C2H4Molecular Formula

C6H6Molecular FormulaCHEmpirical Formula

H2O2Molecular FormulaHOEmpirical FormulaH:O 1:1Smallest Whole Number Ratio

Two compounds can have identical empirical formulas and different molecular formulas.

CalculatingEmpirical Formulas

Step 1 Assume a definite starting quantity (usually 100.0 g) of the compound, if if the actual amount is not given, and express the mass of each element in grams.Step 2 Convert the grams of each element into moles of each element using each elements molar mass.

Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value. If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. If the numbers obtained are not whole numbers, go on to step 4.

Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbersUse these whole numbers as the subscripts in the empirical formula.FeO1.5Fe1 x 2O1.5 x 2Fe2O3

The results of calculations may differ from a whole number.

If they differ 0.1 round off to the next nearest whole number.2.9 3Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

Some Common Fractions and Their Decimal EquivalentsCommon FractionDecimal Equivalent0.3330.250.50.750.666Multiply the decimal equivalent by the number in the denominator of the fraction to get a whole number.

Problems

The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 1 Express each element in grams. Assume 100 grams of compound.K = 56.58 gC = 8.68 gO = 34.73 g

The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 2 Convert the grams of each element to moles.

The analysis of a salt shows that it contains 56.58% potassium (K); 8.68% carbon (C); and 34.73% oxygen (O). Calculate the empirical formula for this substance.Step 3 Divide each number of moles by the smallest value.The simplest ratio of K:C:O is 2:1:3Empirical formula K2CO3

The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Step 1 Express each element in grams. Assume 100 grams of compound.N = 25.94 gO = 74.06 g

Step 2 Convert the grams of each element to moles.The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

Step 3 Divide each number of moles by the smallest value.The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance.

Step 4 Multiply each of the values by 2.The percent composition of a compound is 25.94% nitrogen (N), and 74.06% oxygen (O). Calculate the empirical formula for this substance. Empirical formula N2O5N: (1.000)2 = 2.000O: (2.500)2 = 5.000

Calculating the Molecular Formula from the Empirical Formula

The molecular formula can be calculated from the empirical formula if the molar mass is known.The molecular formula will be equal to the empirical formula or some multiple, n, of it.To determine the molecular formula evaluate n. n is the number of units of the empirical formula contained in the molecular formula.

What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2 g? The molecular formula is (CH2)9 = C9H18Let n = the number of formula units of CH2. Calculate the mass of each CH2 unit 1 C = 1(12.01 g) = 12.01g 2 H = 2(1.01 g) = 2.02g 14.03g