2/9/2015

1

Mass of Part / Mass of Whole

Unit 9: Chemical Formulas

Lesson 9.2: Percent Composition, Empirical & Molecular Formulas

40

Percent Composition

• Sometimes it’s useful to know the composition of a compound in terms of what percentage of the total is each element

• Percent – “Parts per 100”

– The number of specific items per a group of 100 items

– 50% of $100 is $50 (50 items/100 total items)

Percent Example

• You have 4 oranges and 5 apples. What percent of the total is oranges?

• In “parts per 100 (cent)”

oranges 44%100% total9

oranges 4

oranges 44%100% total100

oranges 44

2/9/2015

2

Percent Composition – In Chemistry

• It is the percent by mass of each element in a compound

• Can be determined

– By its chemical formula

– Molar masses of the elements that compose the compound

• The percent of each element contributes to the mass of the compound

100%

compoundtheofmassmolar

elementeachofmasselementeach of mass %

compound in the

Calculating Percent Composition from a Formula

• What is the percent composition of each element in NH4OH?

g14.01 = g14.011:N

g5.04 = g0078.15:H

g16.00 = g00.611:O

g35.05 =

%100g35.05

g14.01 :N

%100g35.05

g5.04 :H

%100g35.05

g16.00 :O

N 39.97%

H 14.38%

O 45.65%

Determine the contribution of each

element

Molar mass

Calculating Percent Composition from a Sample

• Experiments show that the 80.0 grams of a compound is made up of 59.53 grams of oxygen and 20. 47 grams of nitrogen. What is the % composition, by mass, of each element in the compound??

%.

100000 g8

g20.47 :N

%100g 80.00

g59.53 :O

N 25.59%

O 74.41%

Determine the contribution of each

element

2/9/2015

3

Percent Composition What is the mass of nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2.

1. Molar Mass of C6H14N2O2 C = 6(12.01) = 72.06 H =14(1.01) = 14.14

MM = 146.22

28.02 g ___________ x 100% 146.22 g

= 19.16%

3. 0.1916 x 85.0 mg = 16.3 mg N

2.

N = 2(14.01) = 28.02 O = 2(16.00) = 32.00

1. What is the % composition, by mass, of each element in SiO2?

2. Analysis of 35.00 g sample of a substance determined that it contains 18.25 g carbon, 4.60 g hydrogen, and 12.15 g oxygen. What is the % composition, by mass, of each element in the compound?

47

Learning Check!

What is the percent carbon in C5H8NO4 (the

glutamic acid used to make MSG

monosodium glutamate), a compound used

to flavor foods and tenderize meats?

a) 8.22 %C

b) 24.3 %C

c) 41.1 %C

Learning Check!

2/9/2015

4

Chemical Formulas of Compounds

• Formulas give the relative numbers of atoms or

moles of each element in a formula unit - always a

whole number ratio (the law of definite

proportions).

NO2 2 atoms of O for every 1 atom of N

1 mole of NO2 : 2 moles of O atoms to every 1

mole of N atoms

• If we know or can determine the relative number

of moles of each element in a compound, we can determine a formula for the compound.

Do These Make Sense?

50

Coefficients Coefficients tell you how many moles of the compound you have.

So, you have to multiply all atoms in the compound by the coefficient to find the total amount of atoms.

1 is not written as a coefficient.

51

2/9/2015

5

How Do We Determine Formulas?

Qualitative Analysis:

Use a spectrometer to find the atoms in the compound.

Quantitative Analysis:

See how the compound behaves in chemical reactions and use our measurements to figure out the formula

52

Types of Formulas 1. Molecular Formula

The formula that states the actual number of each kind of atom found in one molecule of the compound.

Covalent compounds only!

53

Methane (CH4) Benzene (C6H6)

2. Empirical Formula

The formula of a compound that expresses the smallest whole number ratio of the atoms present.

All ionic compounds & network solids (the “formula unit”)

54

Sodium Chloride (NaCl) Silicon Dioxide (SiO2)

2/9/2015

6

Empirical Formulas of Molecular Compounds

Molecular Compounds can also be stated as empirical formulas.

Some are, normally (Ex. CO2, H2O)

Others aren’t:

55

Methane (CH4) Empirical (CH4)

Ethane (C2H6) Empirical (CH3)

Molecular formula = n х Empirical formula

Determining Empirical Formulas (For Molecular Substances)

• What is the empirical formula for the following compounds?

1. Glucose (C6H12O6)

2. Ethylene Glycol (C2H6O2)

3. Hydrogen peroxide (H2O2)

4. Butane (C4H10)

5. Octane (C8H18)

6. Sucrose (C12H22O11)

7. Glycerol (C3H8O3)

8. Decane (C10H22)

• (CH2O)

• (CH3O)

• (HO)

• (C2H5)

• (C4H9)

• (C12H22O11)

• (C3H8O3)

• (C5H11)

To obtain an Empirical Formula 1. Use percent composition to determine the mass

in grams of each element present – assume you have a 100 g sample.

2. Calculate the number of moles of each element.

3. Divide each number of moles by the smallest one to obtain the simplest whole number ratio.

4. If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers

* Be careful! Do not round off numbers prematurely

2/9/2015

7

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.

require mole ratios so convert grams to moles

moles of N = 2.34g of N = 0.167 moles of N

14.01 g/mole

moles of O = 5.34 g = 0.334 moles of O

16.00 g/mole

Formula: 0.334 0.167ON 0.167 0.334 2

0.167 0.167

N O NO

Calculate Empirical Formula from Percent Composition

• Lactic acid has a molar mass of 90.08 g and has this percent composition:

• 40.0% C, 6.71% H, 53.3% O

• What is the empirical and molecular formula of lactic acid?

• Assume a 100.0 g sample size

– Convert percent numbers to grams

Calculate Empirical Formula from Percent Composition

• Convert mass of each element to moles

• Divide each mole quantity by the smallest number of moles

CmolCg

CmolCg 33.3

0.120.40

HmolHg

HmolHg 66.6

008.171.6

OmolOg

OmolOg 33.3

00.163.53

00.13.33

3.33 :C

00.23.33

6.66 :H

00.13.33

3.33 :O

The ratio of C to H to O is 1 to 2 to 1 Empirical

formula is

Empirical formula mass = 12.01 + 2 (1.008) + 16.00 = 30.03 g/mol

CH2O

2/9/2015

8

Determination of the Molecular Formula

• Obtain the value of n (whole number multiplier)

• Multiply the empirical formula by the multiplier

)/(

)/(

molgmassformulaempirical

molgmassmolarn 3

/03.30

/08.90

molg

molg

Molecular formula = n х empirical formula

Molecular formula = 3 (CH2O) C3H6O3

Relating Empirical and Molecular Formulas

• n represents a whole number multiplier from 1 to as large as necessary

• Calculate the empirical formula and the mass of the empirical formula

• Divide the given molecular mass by the calculated empirical mass – Answer is a whole number multiplier

)/(

)/(

molgmassformulaempirical

molgmassmolarn

Molecular formula = n х Empirical formula

63

What now?

Any Questions?