Percent Composition

Percent Composition• Percent Composition – the percentage by mass of

each element in a compound

Percent = _______PartWhole

x 100%

Percent compositionof a compound or =molecule

Mass of element in 1 mol____________________Mass of 1 mol

x 100%

Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)?

Molar Mass of KMnO4K = 1(39.1) =

39.1Mn = 1(54.9) = 54.9O = 4(16.0) = 64.0

MM = 158 g

Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)? = 158 g

% K

Molar Mass of KMnO4

39.1 g K158 g

x 100 =24.7 %

% Mn54.9 g Mn158 g

x 100 =34.8 %

% O64.0 g O158 g

x 100 = 40.5 %K = 1(39.10) = 39.1

Mn = 1(54.94) = 54.9

O = 4(16.00) = 64.0MM = 158

Percent CompositionDetermine the percentage composition of sodium carbonate (Na2CO3)?

Molar Mass Percent Composition

% Na =46.0 g106 g

x 100% =43.4 %

% C =12.0 g106 g

x 100% =11.3 %

% O =48.0 g106 g

x 100% =45.3 %

Na = 2(23.00) = 46.0C = 1(12.01) = 12.0O = 3(16.00) = 48.0 MM= 106 g

Percent CompositionDetermine the percentage composition of ethanol (C2H5OH)?

% C = 52.13%, % H = 13.15%, % O = 34.72%

_______________________________________________

Determine the percentage composition of sodium oxalate(Na2C2O4)?

% Na = 34.31%, % C = 17.93%, % O = 47.76%

Percent CompositionCalculate the mass of bromine in 50.0 g of Potassium bromide.

1. Molar Mass of KBr

K = 1(39.10) = 39.10 Br =1(79.90) =79.90

MM = 119.0

79.90 g ___________119.0 g

= 0.6714

3. 0.6714 x 50.0g = 33.6 g Br

2.

HydratesHydrated salt – salt that has water molecules trapped within the crystal lattice

Examples: CuSO4•5H2O , CuCl2•2H2O

Anhydrous salt – salt without water molecules

Examples: CuCl2

Can calculate the percentage of water in a hydrated salt.

Percent CompositionCalculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O.

1. Molar Mass of Na2CO3•10H2ONa = 2(22.99) = 45.98

C = 1(12.01) = 12.01

MM = 286.2

H = 20(1.01) = 20.2O = 13(16.00)= 208.00

H = 20(1.01) = 20.2

Water

O = 10(16.00)= 160.00MM = 180.2

2.

3.180.2 g _______286.2 g

62.96 % x 100%=

orH = 2(1.01) = 2.02O = 1(16.00) = 16.00

MM H2O = 18.02

So… 10 H2O = 10(18.02) = 180.2

Percent CompositionCalculate the percentage of water in Aluminum bromide hexahydrate, AlBr3•6H2O.

1. Molar Mass of AlBr3•6H2OAl = 1(26.98) = 26.98

Br = 3(79.90) = 239.7

MM = 374.8

H = 12(1.01) = 12.12O = 6(16.00) = 96.00

H = 12(1.01) = 12.1

Water

O = 6(16.00)= 96.00MM = 108.1

2.

3.108.1 g _______374.8 g

28.85 % x 100%=

orMM = 18.02For 6 H2O = 6(18.02) = 108.2

EMPIRICAL AND MOLECULAR FORMULAS

Formulas

Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms.

Molecular Formula – actual formula of a compound showing the number of atoms present

Percent composition allows you to calculate the simplest ratio among the atoms found in a compound.

Examples:

C4H10 - molecular

C2H5 - empirical

C6H12O6 - molecular

CH2O - empirical

Formulas

Is H2O2 an empirical or molecular formula?

Molecular, it can be reduced to HOHO = empirical formula

Calculating Empirical FormulaAn oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula.

1. Determine the number of grams of each element in the compound.

4.151 g Al and 3.692 g O

2. Convert masses to moles.

4.151 g Al 1 mol Al

26.98 g Al= 0.1539 mol Al

3.692 g O 1 mol O

16.00 g O= 0.2308 mol O

Calculating Empirical FormulaAn oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula.

3. Find ratio by dividing each element by smallest amount of moles.

0.1539 moles Al

0.1539 = 1.000 mol Al

0.2308 moles O

0.1539 = 1.500 mol O

4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)

O = 1.500 x 2 = 3Al = 1.000 x 2 = 2

therefore, Al2O3

Calculating Empirical FormulaA 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound.

4.550 g Co1 mol Co

58.93 g Co= 0.07721 mol Co

5.475 g Cl 1 mol Cl

35.45 g Cl= 0.1544 mol Cl

0.07721 mol Co 0.1544 mol Cl

0.07721 0.07721 = 2= 1

CoCl2

Calculating Empirical FormulaWhen a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula.

2.000 g Fe 1 mol Fe

55.85 g Fe= 0.03581 mol Fe

0.573 g O 1 mol O

16.00 g= 0.03581 mol O

Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g

1 : 1

FeO

Calculating Empirical FormulaA sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

1.3813 g Pb1 mol Pb

207.2 g Pb= 0.006667 mol Pb

0.00672 gH 1 mol H

1.008 g H= 0.00667 mol H

0.4995 g As 1 mol As

74.92 g As= 0.006667 mol As

0.4267g Fe 1 mol O

16.00 g O= 0.02667 mol O

Calculating Empirical FormulaA sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.

0.006667 mol Pb

0.00667 mol H

0.006667 mol As

0.02667 mol O

0.006667

0.006667

0.006667

0.006667

= 1.000 mol Pb

= 1.00 mol H

= 1.000 mol As

= 4.000 mol O

PbHAsO4

Calculating Empirical FormulaThe most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.

Step 1:

In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O.

Step 2:

63.38 g C 1 mol C

12.01 g C= 5.302 mol C

12.38 g N 1 mol N

14.01 g N= 0.8837 mol N

9.80 g H 1 mol H

1.01 g H= 9.72 mol H

14.14 g O 1 mol O

16.00 g O= 0.8832 mol O

Calculating Empirical FormulaThe most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.

Step 3:

5.302 mol C

0.8837= 6.000 mol C

0.8837 mol N

0.8837= 1.000 mol N

9.72 mol H

0.8837= 11.0 mol H

0.8837 mol O

0.8837= 1.000 mol O

6:1:11:1

C6NH11O

Calculating Molecular FormulaA white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula?

Step 1: Molar Mass

P = 2 x 30.97 g = 61.94gO = 5 x 16.00g = 80.00 g

141.94 g

Step 2: Divide MM by Empirical Formula Mass

238.88 g141.94g

= 2

Step 3: Multiply

(P2O5)2 =

P4O10

Calculating Molecular FormulaA compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?

C = 12.01 gH = 1.01 g 13.01 g

78 g/mol

13.01 g/mol= 6

(CH)6 =

C6H6