MOLAR MASS &PERCENT COMPOSITION

MOLES

• How many moles are present?

CH4

C = 1 mol, H = 4 mol.

MOLAR MASS

CH4

Mass of 1 mol of C = 1 x 12.0 g = 12.0 g

Mass of 4 mol of H = 4 x 1.0 g = 4.0 g

Mass of 1 mol of CH4 = 16.0 g

Molar mass of any substance is the mass (in grams) of 1 mol of the substance. The molar mass is obtained by summing the masses of the component atoms.

MOLAR MASS

H2O

Mass of 2 mol of H = 2 x 1.0 g = 2.0 g

Mass of 1 mol of O = 1 x 16.0 g = 16.0 g

Mass of 1 mol of H2O = 18.0 g

PERCENT COMPOSITION

Mass percent = mass element present x 100

mass compound

The % mass that one particular element contributes to an The % mass that one particular element contributes to an entire compound.entire compound.

• to calculate… to calculate…

~~ find total mass of one molecule find total mass of one molecule

~~ ÷÷ mass mass ofof element element (part)(part) byby total mass (whole)total mass (whole)

~ x by 100 ~ x by 100

EXAMPLE• What is the Mass Percent of Na2S?

• Step 1 find Molar Mass.

• Na = 2 x 23.0 = 46.0 g

• S = 1 x 32.0 = 32.0 g

• Molar Mass = 78.0g

• Step 2 divide individual elements mass by the molar mass of entire compound.

Sodium = (46.0 g / 78.0 g) x 100 = 59.0%

Sulfur = (32.0g / 78.0g) x 100 = 41.0%

PRACTICE #1

• Calculate the mass percent of C10H14O.

• Carbon = 10 x 12.0g = 120.0g

• Hydrogen = 14 x 1.0g = 14.0g

• Oxygen = 1 x 16.0g = 16.0g

• Molar mass = 150.0g

PRACTICE #1

Mass Carbon = 120.0g /150.0g x 100 = 80.0%

Mass Hydrogen = 14.0g/150.0g x 100 = 9.3%

Mass Oxygen = 16.0g / 150.0 g x 100 = 10.7%

PRACTICE #2

• Determine mass percent of sulfuric acid.

• H4(SO4)2

• Hydrogen = 4 x 1.0g = 4.0g

• Sulfur = 2 x 32.1g = 64.2g

• Oxygen = 8 x 16.0g = 128.0g

• Molar Mass = 196.2g

PRACTICE #2

Mass Hydrogen = 4.0g / 196.2g x 100 = 2.04%

Mass Sulfur = 64.2g / 196.2g x 100 = 32.7%

Mass Oxygen = 128.0g / 196.2g x 100 = 65.2%

PRACTICE #3

• Determine mass percent of sulfuric acid.

• Ca(NO3)2

• Calcium = 1 x 40.1g = 40.1g

• Nitrogen = 2 x 14.0g = 28.0g

• Oxygen = 6 x 16.0g = 96.0g

• Molar Mass = 164.1g

PRACTICE #3

Mass Calcium = 40.1g / 164.1g x 100 = 24.4%

Mass Nitrogen = 28.0g / 164.1g x 100 = 17.1%

Mass Oxygen = 96.0g / 164.1g x 100 = 58.5%

EMPIRICAL AND MOLECULAR FORMULAS

• Empirical formula---lowest ratio of elements in compound

• Molecular formula ---actual ratio of elements in compounds

• Subscripts are mole ratios

• Compounds may have the same empirical formula but different molecular formulas.

• For example

CH is the empirical formula for C2H2 and C6H6

• If the subscripts can be reduced, then the formula is a molecular formula.

Empirical FormulasEmpirical Formulas

• The lowest whole # ratio of aThe lowest whole # ratio of acompound. (reduce the subscripts down if you can)compound. (reduce the subscripts down if you can)

• Ex. the empirical formula for glucose (CEx. the empirical formula for glucose (C66HH1212OO66) would ) would

be…be… CHCH22OO

• to calculate from % composition…to calculate from % composition…

1. 1. ÷÷ given % by atomic mass given % by atomic mass (to find moles)(to find moles)

2.2. ÷÷ each mole in step 1 by lowest mole value each mole in step 1 by lowest mole value (to find ratio) (to find ratio)

3. 3. values from step 2 = subscripts values from step 2 = subscripts

~~ subscripts subscripts MUSTMUST be a whole # be a whole # (multiply if they are not!)(multiply if they are not!)

Empirical FormulasEmpirical Formulas

• practice…practice…

1.1. What is the empirical formula of a compound thatWhat is the empirical formula of a compound that

contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen?contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen?

NaNa == 36.5%36.5%

23g23g== 1.59 moles1.59 moles

SS == 25.4%25.4% 32g32g

== 0.794 moles0.794 moles

OO == 38.1%38.1%

16g16g== 2.38 moles2.38 moles

******

0.7940.794== 22

0.7940.794 == 11

0.7940.794== 33

subscriptssubscripts

Na SONa SO22 33

Empirical FormulasEmpirical Formulas

• more practice…more practice…

2.2. What is the empirical formula of a compound thatWhat is the empirical formula of a compound that

contains 72.4% iron (Fe) and 27.6% oxygen?contains 72.4% iron (Fe) and 27.6% oxygen?

FeFe == 72.4%72.4%

5656== 1.293 moles1.293 moles

OO == 27.6%27.6%

1616== 1.725 moles1.725 moles

****** 1.2931.293

== 11

1.2931.293== 1.331.33

MUST be a whole #!!!MUST be a whole #!!!

x 3 x 3 == 44

x 3 x 3 == 33

Fe OFe O33 44

FINDING MOLECULAR FORMULAFROM EMPIRICAL FORMULA

• You must be given the actual molecular molar mass of the compound in the problem.

• Divide the actual molecular mass by the empirical formula mass.

• Then multiply all the subscripts by your answer.

• For example the actual molar mass of the previous compound is 88.0g.

• The empirical formula mass is 44.0g

• 88.0g/44.0g = 2

C2H4O X 2 = C4H8O2

LEARNING CHECK!!!WHICH OF THE FOLLOWING ARE EMPIRICAL FORMULAS AND WHICH ARE MOLECULAR?• H2SO4

• C6H12O6

• CH2O

• NaCl

• C6H12

• P4O10

• empirical

• molecular

• empirical

• empirical

• molecular

• molecular