empirical formulas & percent composition

Empirical Formulas & Empirical Formulas & Percent CompositionPercent Composition

CO2 Carbon dioxide

BCl3 boron trichloride

Molar MassMolar Mass

How many How many molesmoles of alcohol of alcohol are there in a “standard” can are there in a “standard” can of beer if there are 21.3 g of of beer if there are 21.3 g of CC22HH66O?O?

(a)Molar mass of C2H6O = 46.08 g/mol

(b)Calc. moles of alcohol

21.3 g • 1 mol

46.08 g = 0.462 mol

How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C2H6O?

= 2.78 x 1023 molecules

We know there are 0.462 mol of C2H6O.

0.462 mol • 6.022 x 1023 molecules

1 mol

How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C2H6O?

= 5.57 x 1023 C atoms

There are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023 molecules • 2 C atoms1 molecule

Empirical & Molecular Empirical & Molecular FormulasFormulas

A pure compound always consists of the same elements combined in the same proportions by weight.

Therefore, we can express molecular composition as PERCENT BY WEIGHT

Ethanol, C2H6O52.13% C13.15% H 34.72% O

Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider some of the family of nitrogen-Consider some of the family of nitrogen-

oxygen compounds:oxygen compounds:

NONO22, nitrogen dioxide, nitrogen dioxide

Structure of NOStructure of NO22

What is the weight percent of N and of O?What is the weight percent of N and of O?

Percent Percent CompositionComposition

Percent Percent CompositionComposition

Consider NOConsider NO22, Molar mass =, Molar mass =

Percent of N and of O?Percent of N and of O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N

46.0 g NO2 • 100% = 30.4 %Wt. % N =

14.0 g N46.0 g NO2

• 100% = 30.4 %

What are percentages of N and O in NO?What are percentages of N and O in NO?

46 g/ mol46 g/ mol

Determining Determining FormulasFormulas

Determining Determining FormulasFormulas

In chemical analysis we determine the % by weight of each element in a given

amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.

PROBLEM: What is its empirical formula of a compound of B and H

if 81.10% is boron?

%BB + %H = 100%

100% - 81.10% B = 18.90% H.If there were 100.0 g of the compound:

there are 81.10 g of B and 18.90 g of H.

Calculate the number of MOLES of each.

Determination of empirical Determination of empirical formulas:formulas:

mass of eachmass of each


mass of eachmass of each

Calculate the number of moles of each element assuming 100.0 g total.

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H


moles of eachmoles of each


moles of eachmoles of each

Take the ratio of moles of B and H.

atoms combine in small whole number ratios


moles ratiomoles ratio


moles ratiomoles ratio

Always divide by the smaller number.

18.75 mol H and 7.502 mol B

7.502 mol 7.502 mol

2.5 mol H to 1.0 mol B

But we need a whole number ratio. .

2.5 mol H or 5 mol H

1.0 mol B 2 mol B

EMPIRICAL FORMULA = B2H5

If the empirical formula is B2H5,

what is its molecular formula?We need to do an

EXPERIMENT to find the MOLAR MASS.

MS gives 53.3 g/molCompare empirical mass

= 26.66 g/unit Find the ratio of these masses.

Molecular Molecular formula = formula = BB44HH1010

2 =

26.66

3.53 2 =

26.66

3.53

Determine the formula of a compound of Sn and I

using the following data.

Determine the formula of a compound of Sn and I

using the following data.

• Reaction of Sn and I2 is done using excess Sn.• Mass of Sn in the beginning = 1.056 g

• Mass of Sn remaining

= 0.601 g

• Mass of iodine (I2) used

= 1.947 g

Find the mass of Sn that combined with 1.947 g I2.

Mass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 g

Find moles of Sn used:

0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •

1 mol118.7 g

= 3.83 x 10-3 mol Sn

Tin and Iodine Tin and Iodine CompoundCompound


Now find the number of moles of I2 that combined with 3.83 x 10-3 mol Sn. Mass of I2 used was 1.947 g.

1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2

How many mol of iodine atoms? ?

= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2

7.671 x 10-3 mol I2 2 mol I atoms

1 mol I2


Now find the ratio of number of moles of moles of I and Sn that combined.

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

Empirical formula is Empirical formula is

SnISnI44

Practice Formula Determination

Homework:Worksheets pgs 17-20

Pg 13-14 good review for mole calculations

Prelab questions

empirical formulas & percent composition

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