THE MOLE NOTES PART 2Percent Comp

Empirical FormulasMolecular Formulas

PERCENT COMP: The percent by mass of each element in a

compound

Percent – a part over the whole x 100 Part

x 100Whole

PERCENT COMP: The mass of each element in a compound

divided by the entire mass of the compound then multiplied by 100..

Part = the mass of the element you are wanting the % mass for.

Whole = the molar mass of the entire compound

PERCENT COMP: If we have Mg3(PO4)2 and we want to know the mass

% of Magnesium:

How many magnesium atoms are there in 1 mol? _____What is the molar mass of just Mg? 3 x 24g = 72g of Mg in Mg3(PO4)2

Part: 72g MgWhole: what is the molar mass of the whole compound?72 + (31 x 2) + (16 x 8) = 263g/molPart/whole = 72/263 x 100 = 27%

PERCENT COMP: Now lets calculate the Mass percent of

Phosphorous and Oxygen in Mg3(PO4)2:

Phosphorous:2 phosphorous = 2 x 31 = 6262g/263g x 100 = 23.9% = 24%

Oxygen:8 oxygens= 8 x 16 = 128128g/263g x 100 = 48.7% = 49%

HINT: HOW TO CHECK YOUR WORK

To check your work add up all the Percent Masses of each element:

Mg – 27% P – 24% 100 O – 49%

It should be right around 100…within a number or 2 (this is because we are rounding to the nearest whole number)

PRACTICE: Now complete the following problems on

your Notes handout:

1. Lithium Sulfate

2. Iron (III) Oxide

EMPIRICAL AND MOLECULAR FORMULAS: Empirical formula: the formula for a

compound with the smallest whole number ratio of elements

Molecular formula: The formula that specifies the actual number of atoms of each element in a substance (may or may not be the same as the empirical)

CH4 is the empirical formula for: CH4, C2H8 C3H12 (these are molecular

formulas)

EMPIRICAL FORMULAS:1. Determine the number of moles of each

substance present in the compound. (Use percent comp)

*This will typically be given to you.2. Determine the grams

*Assume that there is a hundred grams of compound, therefore, the percentage is the same as the number of grams. 3. Convert the grams to moles4. Determine the ratio5. Write the Empirical Formula

EMPIRICAL FORMULAS: STEPS 1 AND 21. A compound has 40% sulfur and 60% oxygen.2. How many grams of each?

Sulfur = 40% = 40gOxygen = 60% = 60g

EMPIRICAL FORMULAS: STEP 3 Convert grams to moles: (S=40g and O=60g)

40 g S 1 mol S = 1.25 moles S 32 g S

60 g O 1 mol O = 3.75 moles O 16 g O

EMPIRICAL FORMULAS: STEP 4 Determine the ratio

1.25 moles S and 3.75 moles O

1. Choose the smallest number (1.25 moles S)2. Divide all by the smallest number

1.25 moles S = 1 mol S 3.75 moles O = 3 mol O 1.25 1.25

EMPIRICAL FROMULA: STEP 5 You came up: 1 mol S & 3 mols O

You use these numbers to write the empirical formula:

SO3

PRACTICE: Write the empirical formulas for the following

problem. Complete this on your notes handout.

Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 49% carbon, 8% hydrogen, and 43% oxygen.

MOLECULAR FORMULA:1. Find the empirical formula2. Find the molar mass of the empirical

formula3. Divide the given molar mass by the EF’s

molar mass4. Multiply each subscript of the EF by the

number you go in Step 35. Write the formula

MOLECULAR FORMULA: STEP 1 Find the empirical formula: A compound has the following composition: 77% carbon, 12%

hydrogen, and 11% oxygen. Its molar mass is 282 g/mol, what is its molecular formula.

Assume there are 100g.77g C 1 mol C = 6.42 mols C

12g C12 g H 1 mol H = 12 mols H 1 g H11 g O 1 mol O = .688 mols O 16 g O

MOLECULAR FORMULA: STEP 1 CONT. Find the empirical formula6.42 mols C= 9 12 mols H = 17 .688 mols O = 1.688 mols O .688 mols O .688 mols O

EF: C9H17O

MOLECULAR FORMULA: STEP 2 Find the molar mass of the EF:EF: C9H17OCarbon: (9 x 12) = 108Hydrogen: (17 x 1) = 17Oxygen: (1 x 16) = 16

108 + 17 + 16 = 142g/mol

MOLECULAR FORMULA: STEP 3 Divide the molar mass of the given by the

molar mass of EF.

Given Molar Mass: 282 g/molEF Molar Mass: 142g/mol

282 ÷ 142 = 1.98 (rounds to 2)

MOLECULAR FORMULA: STEP 4 Multiply the subscripts of the EF by the number

you came up with in step 3.

EF: C9H17OAnswer from step 3: 2

C9: 9 x 2 = 18H17: 17 x 2 = 34O: 1 x 2 = 2*these now become the subscripts for the molecular formula.

Molecular Formula: C18H34O2

PRACTICE: MOLECULAR FORMULA Complete the following problem on your

notes problem page.

A colorless liquid composed of 47% nitrogen and 53% oxygen, has a molar mass of 60 g/mol. What is the empirical and molecular formula?