Molar Mass and Percent Composition

Law of Definite Proportions

• John Dalton proposed that regardless of sample size a compound is always made up of elements combined in the same proportion by mass

• This law allows us to determine the mass of a compound from the masses of its component elements.

• It also allows us to determine the percentage of every element in the compound.

• This law allows us to have generic/store brand medications. Although the name seems different, the essential compounds in the medication are chemically the same.

• For example: the brand name of Tylenol ™ is the same as the generic brand of acetaminophen. There are numerous other brand names!

• Acetaminophen is C8H9NO2

Atomic Mass - Element

• The mass of an element is found on the periodic table.

• We will round all masses off the table to two decimal places before using in any calculations!

Compound Mass – Counting Atoms

• For the mass of a compound, you have to consider both the number and types of atoms!

• A subscript (number appearing below) indicates the number of atoms (if molecule) or ions (if formula unit)

Example

• CH4 (methane) contains 1 atom of carbon and 4 atoms of hydrogen

• AlPO4 (aluminum phosphate)contains 1 ion of aluminum and 1 ion of phosphate but it also contains 1 atom of aluminum, 1 atom of phosphorus and 4 atoms of oxygen. We need to use the numbers of atoms in our calculation of molar mass.

Compound Mass – Counting Atoms

• If subscripts appear outside of parentheses, you need to multiply!

• Example: Mg(NO3)2 contains:

1 atom Magnesium

2 atoms of Nitrogen(2 x 1)

6 atoms of Oxygen (2 x 3)

Compound Mass – Counting Atoms

• If you have a hydrate (crystal with water enclosed) the formula will look something like this: CuSO4

. 5H2O

• You need to multiply the number in front of the water molecule to get the correct # of atoms:

Copper – 1 atom, Sulfur – 1atom, Oxygen – 4 atoms, water – 5 molecules or

Cu -1, S – 1, O - (4 + 5) = 9atoms, H – 10atoms

(The dot indicates addition or contains, not multiplication!)

Mass Calculation - Compound

• Multiply the # of atoms of each element in the compound by its corresponding mass from the periodic table (round to two decimal places)

• Round the product to the same # of significant digits as the mass

• Add the products and round based on rule – least # of decimal places

• Unit is amu

Examples!• BaCl2 ( 1 atom Ba x 137.33 ) + (2 atoms Cl x

35.45 ) = 137.33 + 70.90 = 208.23 amu

• Mg(NO3)2 (1 atom Mg x 24.30) + ((2x1) atoms nitrogen x 14.01) + ((2x3) atoms oxygen x 16.00) = 24.30 + 28.02 + 96.00 = 148.32 amu

• NOTE: I am typing these across the page, but on your paper it should be written down the page! See example on the board!

Molar Mass - Element(Text Reference Chapter 11)

The mass of one mole of an element is equivalent to its atomic mass.

Example: one atom of sodium has a mass of 22.99 amu;

22.99 grams Na = 1 mole Na Molar mass is the mass of one mole of

any substance. Molar Mass Sodium = 22.99 g/mol

Molar Mass - Compound

• The mass of one mole of a compound is also based on the number and type of atoms.

• The calculation is the same as the mass of a compound!

• The unit becomes g/mol

Examples!

• BaCl2 ( 1 atom Ba x 137.33 ) + (2 atoms Cl x 35.45 ) = 137.33 + 70.90 = 208.23 g/mol

• Mg(NO3)2 (1 atom Mg x 24.31) + ((2x1) atoms nitrogen x 14.01) + ((2x3) atoms oxygen x 16.00) = 24.31 + 28.02 + 96.00 = 148.33 g/mol

Hydrates

• Learn the molar mass of water! H2O (2 atoms Hydrogen x 1.01) + (1atom oxygen x 16.00) = 2.02 + 16.00 = 18.02 g/mol

• CuSO4 . 5H2O (1 atom Cu x 63.55) + (1 atom Sulfur x 32.06) + (4 atoms Oxygen x 16.00) + (5 molecules of water x 18.02) = 63.55 + 32.06 + 64.00 + 90.10 = 249.71 g/mol

Molar Mass Practice!

Calculate the molar mass of:

1. Sodium chloride, NaCl

Continued Practice

2. Lithium Phosphate, Li3PO4

Continued Practice

3. Manganese (VII) carbonate, Mn2(CO3)2

Continued practice

4. Ferric chloride hexahydrate, FeCl3 . 6H2O

Check it!

1.) NaCl (1 x 22.99) + ( 1 x 35.45) = 58.44 g/mol

2. )Li3PO4 (3 x 6.94) + (1 x 30.97) + (4 x 16.00) = 20.8 +30.97 + 64.00 = 115.77 115.8g/mol

3. )Mn2(CO3)2 (2 x 54.94) + (2 x 12.01) + (6 x 16.00) = 109.9 + 24.02 + 96.00 = 229.92 229.9 g/mol

4.) FeCl3 . 6H2O (1 x 55.85) + (3 x 35.45) + (6 x

18.02) = 55.85 + 106.4 + 108.1 = 270.35 270.4 g/mol

Li3PO4

Mole Conversions

• 1 mole = molar mass (g)

• This equivalent relationship can be used to convert between moles and grams and reverse.

• Must calculate the molar mass first if a compound.

Example:

• How many moles are in 65.0 grams of sodium chloride?

• Given: 65.0g NaCl• Want: moles NaCl• Relationship – molar mass NaCl (calculated

before): 58.44 g/mol • Means 58.44 g NaCl = 1 mol NaCl

• Possible conversion factors:

58.44 g NaCl/1mol NaCl or

1mol NaCl/58.44g NaCl

• Set up:

65.0g NaCl x 1mol NaCl = 1.11mol NaCl

58.44 g NaCl

Example:

• How many grams are in 1.25 moles of lithium phosphate?

• Given: 1.25 moles Li3PO4

• Want: grams Li3PO4

• Relationship: molar mass Li3PO4 (calculated earlier): 115.8g/mol

• Means 115.8 g Li3PO4 = 1mole Li3PO4

• Possible conversion factors:

115.8 g Li3PO4 /1mole Li3PO4 or

1mole Li3PO4 / 115.8 g Li3PO4

Set up:1.25 moles Li3PO4 x 115.8 g Li3PO4 = 145g Li3PO4

1mole Li3PO4

Practice!

5. How many moles are in 3.4g Mg(OH)2?

Continued Practice

6. How many grams are in 0.00500 moles of Ca3N2?

Answers:

5. Need molar mass of Mg(OH)2: (1 x 24.30) + ((2x1)x16.00) + ((2 x 1) x 1.01) = 24.30 + 32.00 + 2.02 = 58.32 g/mol

Means: 58.32 g Mg(OH)2 = 1mol Mg(OH)2

3.4 g Mg(OH)2 x 1mole = 0.058mol

58.32 g

Answers (cont.)

6. Need molar mass of Ca3N2 : (3 x 40.08) + (2 x 14.01) = 120.2 + 28.02 = 148.2 g/mol

0.00500 moles Ca3N2 x 148.2 g = 0.741g

1 mole

Multiple Conversions

Note: use previously learned conversion factors:

1 mole = 6.02 x 1023 particles ,

1 mole = molar mass(g)

7. How many grams are in 5.75 x 1019

formula units of NaCl?

Continued Practice

8. How many molecules of nitrogen, N2, are in 165 grams of the gas?

Answers:

7. Need molar mass of NaCl: 22.99 + 35.45 = 58.44 g/mol

5.75 x 1019 f.unit NaCl x 1 mole x 58.44 g

6.02 x 1023 1 mole

formula units

= 5.58 x 10-3 g

Answers (cont.)

8. Need the molar mass of N2: 2 x 14.01 = 28.02 g/mol

165 g N2 x 1 mole x 6.02 x 1023 molecules

28.02 g 1 mole

= 3.54 x 1024 molecules

Percent by Mass/Percent Composition of Compound

• You can find the mass of any element in a compound by first calculating the mass and using the general relationship: part/whole x 100 = %. The part is the element mass, the whole is the compound mass.

• If you find the percent of every element you are finding the percent composition.

Example

BaCl2 ( 1 atom Ba x 137.33 ) + (2 atoms Cl x 35.45 ) = 137.33 + 70.90 = 208.23 amu

Percent Composition:

137.33/208.23 x 100 = 65.951% Ba

70.90/208.23 x 100 = 34.05% Cl

Example

Mn2(CO3)2 (2 x 54.94) + (2 x 12.01) + (6 x 16.00) = 109.9 + 24.02 + 96.00 = 229.92 229.9 g/mol

What is the percent by mass of manganese in the above compound?

109.0gMn/229.9g Mn2(CO3)2 x 100 =

47.80% Mn

Practice!

9. Find the percent composition of Acetaminophen: C8H9NO2

Continued practice

10. Find the percent by mass of iron in Iron(III)sulfate: Fe2(SO4)3 .

Answers:

9. ) C8H9NO2

(8 x 12.01) + (9 x 1.01) + (1 x 14.01) + (2 x 16.00) = 96.08gC + 9.09gH + 14.01gN +

32.00gO = 151.18g C8H9NO2

96.08/ 151.18 x 100 = 63.29%C9.09/ 151.18 x 100 = 6.01%H14.01/ 151.18 x 100 = 9.267%N32.00/ 151.18 x 100 = 21.17%O

10. )Fe2(SO4)3

(2 x 55.85) + ((3x1) x 32.06) + ((3 x 4) x 16.00) = 111.7gFe + 96.18gS+ 192.0gO =

399.88 399.9g Fe2(SO4)3

111.7/399.9 x 100 = 27.93%Fe

Hydrates - % Water

• You can calculate the percent of water in the hydrate: CuSO4 . 5H2O (1 atom Cu x 63.55) + (1 atom Sulfur x 32.06) + (4 atoms Oxygen x 16.00) + (5 molecules of water x 18.02) = 63.55 + 32.06 + 64.00 + 90.10 gH2O = 249.71 g/mol CuSO4 . 5H2O

90.10/249.71 x 100 = 36.08% H2O

Practice

11. Determine the percent of water in sodium sulfate decahydrate (Na2SO4 . 10H2O)

Check Answer:

11.) Na2SO4 . 10H2O

(2 x 22.99) + (1 x 32.06) + (4 x 16.00) + (10 x 18.02) = 95.98 + 32.06 + 64.00 + 180.2 = 372.24 = 372.2 g/mol(1 decimal place)

180.2/372.2 x 100 = 48.41%H2O