An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. The percentages of magnesium and oxygen in the compound are calculated to be 60.3% and 39.7% respectively, totaling 100%. Another example calculates the percentages of carbon (81.8%) and hydrogen (18.2%) in propane. The document then provides instructions and examples for calculating empirical formulas based on the percentages of elements in compounds.

1. Terms to KnowTerms to Know
 Percent composition – relative amounts ofPercent composition – relative amounts of
each element in a compoundeach element in a compound
 Empirical formula – lowest whole- numberEmpirical formula – lowest whole- number
ratio of the atoms of an element in aratio of the atoms of an element in a
compoundcompound

2. An 8.20 g piece of magnesiumAn 8.20 g piece of magnesium
combines completely with 5.40combines completely with 5.40
g of oxygen to form ag of oxygen to form a
compound. What is the percentcompound. What is the percent
composition of this compound?composition of this compound?
1. Calculate the total mass1. Calculate the total mass
2.2. Divide each given by the total massDivide each given by the total mass
and then multiply by 100%and then multiply by 100%
3.3. Check your answer: TheCheck your answer: The
percentages should total 100%percentages should total 100%

3. AnswerAnswer
 The total mass is 8.20 g + 5.40 g = 13.60The total mass is 8.20 g + 5.40 g = 13.60
gg
 Divide 8.2 g by 13.6 g and then multiply byDivide 8.2 g by 13.6 g and then multiply by
100% = 60.29412 = 60.3%100% = 60.29412 = 60.3%
 Divide 5.4 g by 13.6 g and then multiply byDivide 5.4 g by 13.6 g and then multiply by
100% = 39.70588 = 39.7%100% = 39.70588 = 39.7%
 Check your answer: 60.3% + 39.7% =Check your answer: 60.3% + 39.7% =
100%100%

4. Calculate the percent compositionCalculate the percent composition
of propane (Cof propane (C33HH88))
 1. List the elements1. List the elements
 2. Count the atoms2. Count the atoms
 3. Multiply the number of atoms of3. Multiply the number of atoms of
the element by the atomic mass ofthe element by the atomic mass of
the element (atomic mass is on thethe element (atomic mass is on the
periodic table)periodic table)
 4. Express each element as a4. Express each element as a
percentage of the total molar masspercentage of the total molar mass
 5. Check your answer5. Check your answer

5. 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%

6. Calculate the mass of carbon inCalculate the mass of carbon in
52.0 g of propane (C52.0 g of propane (C33HH88))
1.1. Calculate the percent composition usingCalculate the percent composition using
the formula (See previous problem)the formula (See previous problem)
2. Determine 81.8% of 82.0 g2. Determine 81.8% of 82.0 g
Move decimal two places to theMove decimal two places to the
left (.818 x 82 g)left (.818 x 82 g)
3. Answer = 67.1 g3. Answer = 67.1 g

7. Calculating Empirical FormulasCalculating Empirical Formulas
 Microscopic – atomsMicroscopic – atoms
 Macroscopic – moles of atomsMacroscopic – moles of atoms
 Lowest whole-number ratio may not be theLowest whole-number ratio may not be the
same as the compound formulasame as the compound formula
Example: The empirical formula ofExample: The empirical formula of
hydrogen peroxide (Hhydrogen peroxide (H22OO22) is HO) is HO

8. Empirical FormulasEmpirical Formulas
 The first step is to find the mole-to-moleThe 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 elementsthese will be the subscripts of the elements
in the formulain 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 Cand C88HH88 have an empiricalhave an empirical
formula of CHformula of CH
 If either or both numbers are not wholeIf either or both numbers are not whole
numbers, numbers in the ratio must benumbers, numbers in the ratio must be
multiplied by the same number to yieldmultiplied by the same number to yield
whole number subscriptswhole number subscripts

9. What is the empirical formula ofWhat is the empirical formula of
a compound that is 25.9%a compound that is 25.9%
nitrogen and 74.1% oxygen?nitrogen and 74.1% oxygen?
 1. Assume 100 g of the compound, so that1. Assume 100 g of the compound, so that
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 moleDivide 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 the3. Divide both molar quantities by the
smaller number of molessmaller number of moles

10.  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 each5. Multiply by a number that converts each
to a whole number (In this case, theto a whole number (In this case, the
number is 2 because 2 x 2.5 = 5, which isnumber is 2 because 2 x 2.5 = 5, which is
the smallest whole number )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

11. Determine the Empirical FormulasDetermine the Empirical Formulas
 1. H1. H22OO22
 2. CO2. CO22
 3. N3. N22HH44
 4. C4. C66HH1212OO66
 5. What is the empirical formula of a5. What is the empirical formula of a
compound that is 3.7% H, 44.4% C, andcompound that is 3.7% H, 44.4% C, and
51.9% N?51.9% N?

12. AnswersAnswers
 Compound Empirical FormulaCompound Empirical Formula
 1. H1. H22OO22 HOHO
 2. CO2. CO22 COCO22
 3. N3. N22HH44 NHNH22
 4. C4. C66HH1212OO66 CHCH22OO
 5. HCN5. HCN

13. Calculating Molecular FormulasCalculating Molecular Formulas
 The molar mass of a compound is aThe molar mass of a compound is a
simple whole-number multiple of thesimple whole-number multiple of the
molar mass of the empirical formulamolar mass of the empirical formula
 The molecular formula may or mayThe molecular formula may or may
not be the same as the empiricalnot be the same as the empirical
formulaformula

14. Calculate the molecular formulaCalculate the molecular formula
of the compound whose molarof the compound whose molar
mass is 60.0 g and empiricalmass is 60.0 g and empirical
formula is CHformula is CH44N.N.
 1. Using the empirical formula, calculate the1. Using the empirical formula, calculate the
empirical formula mass (efm)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 efm2. Divide the known molar mass by the efm
 3. Multiply the formula subscripts by this value3. Multiply the formula subscripts by this value
to get the molecular formulato get the molecular formula

15. 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

16. Practice ProblemsPractice Problems
 1) What is the empirical formula of a compounds1) What is the empirical formula of a compounds
that is 25.9% nitrogen and 74.1% oxygen?that is 25.9% nitrogen and 74.1% oxygen?
2) Calculate the empirical formula of a compound2) Calculate the empirical formula of a compound
that is 32.00% C, 42.66% O, 18.67% N, and 6.67%that is 32.00% C, 42.66% O, 18.67% N, and 6.67%
H.H.
3) Calculate the empirical formula of a compound3) Calculate the empirical formula of a compound
that is 42.9% C and 57.1% O.that is 42.9% C and 57.1% O.

17. Practice ProblemsPractice Problems
 4) What is the molecular formula for each compound:4) What is the molecular formula for each compound:
a) CHa) CH22O: 90 gO: 90 g
b) HgCl: 472.2 gb) HgCl: 472.2 g
c) Cc) C33HH55OO22: 146 g: 146 g