Terms to KnowTerms to Know

Percent composition – relative Percent composition – relative amounts of each element in a amounts of each element in a compoundcompound

Empirical formula – lowest whole- Empirical formula – lowest whole- number ratio of the atoms of an number ratio of the atoms of an element in a compoundelement in a compound

An 8.20 g piece of An 8.20 g piece of magnesium combines magnesium combines

completely with 5.40 g of completely with 5.40 g of oxygen to form a oxygen to form a

compound. What is the compound. What is the percent composition of this percent composition of this

compound?compound?1. Calculate the total mass1. Calculate the total mass

2.2. Divide each given by the total Divide each given by the total mass mass

and then multiply by 100%and then multiply by 100%

3.3. Check your answer: The Check your answer: The

percentages should total 100%percentages should total 100%

AnswerAnswer

The total mass is 8.20 g + 5.40 g = The total mass is 8.20 g + 5.40 g = 13.60 g13.60 g

Divide 8.2 g by 13.6 g and then multiply Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3%by 100% = 60.29412 = 60.3%

Divide 5.4 g by 13.6 g and then multiply Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7%by 100% = 39.70588 = 39.7%

Check your answer: 60.3% + 39.7% = Check your answer: 60.3% + 39.7% = 100%100%

Calculate the percent Calculate the percent composition of propane (Ccomposition of propane (C33HH88))

1. List the elements1. List the elements 2. Count the atoms2. Count the atoms 3. Multiply the number of atoms 3. Multiply the number of atoms

of the element by the atomic of the element by the atomic mass of the element (atomic mass of the element (atomic mass is on the periodic table)mass is on the periodic table)

4. Express each element as a 4. Express each element as a percentage of the total molar percentage of the total molar massmass

5. Check your answer5. Check your answer

AnswerAnswer

Total molar mass = 44.0 g/molTotal molar mass = 44.0 g/mol 36.0 g C = 81.8%36.0 g C = 81.8% 8.0 g H = 18.2%8.0 g H = 18.2%

Calculate the mass of carbon Calculate the mass of carbon in 52.0 g of propane (Cin 52.0 g of propane (C33HH88))

1.1. Calculate the percent composition Calculate the percent composition using the formula (See previous using the formula (See previous problem)problem)

2. Determine 81.8% of 82.0 g2. Determine 81.8% of 82.0 g

Move decimal two places to the Move decimal two places to the

left (.818 x 82 g)left (.818 x 82 g)

3. Answer = 67.1 g3. Answer = 67.1 g

Calculating Empirical Calculating Empirical FormulasFormulas

Microscopic – atomsMicroscopic – atoms Macroscopic – moles of atomsMacroscopic – moles of atoms Lowest whole-number ratio may not Lowest whole-number ratio may not

be the same as the compound be the same as the compound formula formula

Example: The empirical formula of Example: The empirical formula of hydrogen peroxide (Hhydrogen peroxide (H22OO22) is HO) is HO

Empirical FormulasEmpirical Formulas The first step is to find the mole-to-mole The first step is to find the mole-to-mole

ratio of the elements in the compoundratio of the elements in the compound If the numbers are both whole numbers, If the numbers are both whole numbers,

these will be the subscripts of the these will be the subscripts of the elements in the formulaelements in the formula

If the whole numbers are identical, If the whole numbers are identical, substitute the number 1substitute the number 1

Example: CExample: C22HH22 and C and C88HH88 have an have an empirical formula of CHempirical formula of CH

If either or both numbers are not whole If either or both numbers are not whole numbers, numbers in the ratio must be numbers, numbers in the ratio must be multiplied by the same number to yield multiplied by the same number to yield whole number subscripts whole number subscripts

What is the empirical What is the empirical formula of a compound that formula of a compound that is 25.9% nitrogen and 74.1% is 25.9% nitrogen and 74.1%

oxygen?oxygen? 1. Assume 100 g of the compound, so 1. Assume 100 g of the compound, so

thatthat

there are 25.9 g N and 74.1 g Othere are 25.9 g N and 74.1 g O 2. Convert to mole-to-mole ratio:2. Convert to mole-to-mole ratio:

Divide each by mass of one mole Divide each by mass of one mole

25.9 g divided by 14.0 g = 1.85 mol N25.9 g divided by 14.0 g = 1.85 mol N

74.1 g divided by 16.0 g = 4.63 mol O74.1 g divided by 16.0 g = 4.63 mol O 3. Divide both molar quantities by the 3. Divide both molar quantities by the

smaller number of molessmaller number of moles

4. 1.85/1.85 = 1 mol N4. 1.85/1.85 = 1 mol N

4.63/1.85 = 2.5 mol O4.63/1.85 = 2.5 mol O 5. Multiply by a number that converts 5. Multiply by a number that converts

each to a whole number (In this case, each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number )which is the smallest whole number )

2 x 1 mol N = 22 x 1 mol N = 2 2 x 2.5 mol O = 52 x 2.5 mol O = 5 Answer: The empirical formula is NAnswer: The empirical formula is N22OO55

Determine the Empirical Determine the Empirical FormulasFormulas

1. H1. H22OO22

2. CO2. CO2 2

3. N3. N22HH44

4. C4. C66HH1212OO66

5. What is the empirical formula of a 5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, compound that is 3.7% H, 44.4% C, and 51.9% N?and 51.9% N?

AnswersAnswers

Compound Empirical FormulaCompound Empirical Formula 1. H1. H22OO22 HO HO

2. CO2. CO2 2 COCO2 2

3. N3. N22HH4 4 NHNH22

4. C4. C66HH1212OO66 CH CH22OO

5. HCN5. HCN

Calculating Molecular Calculating Molecular FormulasFormulas

The molar mass of a compound The molar mass of a compound is a simple whole-number is a simple whole-number multiple of the molar mass of the multiple of the molar mass of the empirical formulaempirical formula

The molecular formula may or The molecular formula may or may not be the same as the may not be the same as the empirical formula empirical formula

Calculate the molecular Calculate the molecular formula of the compound formula of the compound

whose molar mass is 60.0 g whose molar mass is 60.0 g and empirical formula is and empirical formula is

CHCH44N.N. 1. Using the empirical formula, calculate 1. Using the empirical formula, calculate

the empirical formula mass (efm)the empirical formula mass (efm) (Use the same procedure used to calculate (Use the same procedure used to calculate

molar mass.) molar mass.) 2. Divide the known molar mass by the 2. Divide the known molar mass by the

efm efm 3. Multiply the formula subscripts by this 3. Multiply the formula subscripts by this

value to get the molecular formulavalue to get the molecular formula

AnswerAnswer

Molar mass (efm) is 30.0 gMolar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 260.0 g divided by 30.0 g = 2 Answer: CAnswer: C22HH88NN22

Practice ProblemsPractice Problems 1) What is the empirical formula of a 1) What is the empirical formula of a

compounds that is 25.9% nitrogen and 74.1% compounds that is 25.9% nitrogen and 74.1% oxygen?oxygen?

2) Calculate the empirical formula of a 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H.18.67% N, and 6.67% H.

3) Calculate the empirical formula of a 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.compound that is 42.9% C and 57.1% O.

Practice ProblemsPractice Problems 4) What is the molecular formula for each 4) What is the molecular formula for each

compound:compound:

a) CHa) CH22O: 90 gO: 90 g

b) HgCl: 472.2 gb) HgCl: 472.2 g

c) Cc) C33HH55OO22: 146 g: 146 g