1. Molar Mass & Percent Composition

2. Moles
How many moles are present?
CH4
C = 1 mol, H = 4 mol.

3. Molar Mass Mass of 4 mol of H = 4 x 1.0 g = 4.0 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.
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 = g
Mass of 1 mol of CH4 = 16.0 g

4. 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

5. Percent Composition
The % mass that one particular element contributes to an entire compound.
to calculate…
~ find total mass of one molecule
~ ÷ mass of element (part) by total mass (whole)
~ x by 100
Mass percent = mass element present x 100 mass compound

6. What is the Mass Percent of Na2S? Step 1 find Molar Mass.
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 = 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%

7. Calculate the mass percent of C10H14O. Carbon = 10 x 12.0g = 120.0g
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

8. 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 / g x 100 = 10.7%

9. 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

10. 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%

11. 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

12. 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%

13. EMPIRICAL AND MOLECULAR FORMULAS
Empirical formula---lowest ratio of elements in compound
Molecular formula ---actual ratio of elements in compounds
Subscripts are mole ratios

14. 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.

15. Empirical Formulas CH2O The lowest whole # ratio of a
compound. (reduce the subscripts down if you can)
Ex. the empirical formula for glucose (C6H12O6) would be…
CH2O
to calculate from % composition…
1. ÷ given % by atomic mass (to find moles)
2. ÷ each mole in step 1 by lowest mole value (to find ratio)
3. values from step 2 = subscripts
~ subscripts MUST be a whole #
(multiply if they are not!)

16. Empirical Formulas Na SO *** practice…
1. What is the empirical formula of a compound that
contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen?
Na = 36.5%
= 1.59 moles
= 2
23g
0.794
Na SO 2 3
S = 25.4%
= moles
***
= 1
32g
subscripts
0.794
O = 38.1%
= 2.38 moles
= 3
16g
0.794

17. Empirical Formulas x 3 = 3 x 3 = 4 Fe O *** more practice…
2. What is the empirical formula of a compound that
contains 72.4% iron (Fe) and 27.6% oxygen?
Fe = 72.4%
= moles
***
= 1
x 3 = 3 56
1.293
O = 27.6%
= moles
= 1.33
x 3 = 4 16
1.293
MUST be a whole #!!!
Fe O 3 4

18. FINDING MOLECULAR FORMULA FROM 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

19. 88.0g/44.0g = 2
C2H4O X 2 = C4H8O2

20. LEARNING CHECK!!! Which of the following are empirical formulas and which are molecular?
H2SO4
C6H12O6
CH2O
NaCl
C6H12
P4O10
empirical
molecular