Find the % Composition of Each Element in the Following Compound Ca3(PO4)2

Chemistry

CH 7 Notes

% Composition or Mass %

Find the % composition of each element in the following compound Ca3(PO4)2

1.  Find compounds formula mass.

Ca 3(40.1) =120.3

P 2(31.0) = 62.0

O 8(16.0) = 128.0

310.3 formula mass

2. Divide the total formula mass into the mass of each element and multiply by 100 to change into a %

Ca 3(40.1) =120.3/310.3 = 0.388 x 100 = 38.8% Ca

P 2(31.0) = 62.0/310.3 = 0.200 x 100 = 20.0% P

O 8(16.0) = 128.0/310.3 = 0.413 x 100 = 41.3% O

310.3 formula mass

Find the % by mass of water in the following compound CuSO4* 5 H2O

1.  Find compounds formula mass.

Cu 1(63.5) = 63.5

S 1(32.1) = 32.1

O 4(16.0) = 64.0

H2O (2 x 1.0 + 16= 18.0) 5(18.0) = 90.0

249.6

Take waters total and divide by formula mass 90.0/249.6 = 0.361 x 100 = 36.1% water

Empirical Formula-(is a reduced molecular formula) C3H9 molecular; CH3 empirical

Find the empirical formula of a compound that is 32.38% Na, 22.65% S and 44.9% O. Since the percentages add to 100, you can change them directly into grams, thus 32.38g Na, 22.65g S and 44.9g O.

1.  Find moles of each element

32.38g Na 1 moles = 1.41 n

23.0g

22.65g S 1 moles = 0.71 n

32.1g

44.90g O 1 moles = 2.81 n

16.0g

2. Find mole ratio by dividing all moles by the smallest mole

1.41n Na/0.71 = 1.985 close enough to 2

0.71n S/ 0.71 = 1

2.81n O/0.71 = 3.957 close enough to 4

3. Insert ratios and subscripts Na2SO4

4. If the ratio come out as 1:1.5:3 double all numbers; thus 2:3:6 and insert as subscripts.

Molecular Formulas

1.  Need the empirical formula. Either it will be given in the problem or you will have to calculate it as in the above example.

2.  Find the formula mass of the empirical formula and divide that into the molecular formulas mass given in the problem to see how many times heavier the molecular is than the empirical.

3.  Then multiply that number into the empirical formula to find molecular formula.

A compound with a molecular mass of 42.08amu is found to be 85.64% carbon and 14.36% hydrogen by mass. Find the molecular formula.

1. Find empirical formula:

85.64g C 1 mole = 7.14n C/7.14 = 1

12.0g thus empirical is CH2

14.36g H 1 mole = 14.36n H/ 7.14 = 2

1.0g

2. Find formula mass of empirical CH2 C 1(12.0)= 12.0

H 2(1.0) = 2.0

14.0

3. Divide givens molecular mass by empirical mass.

42.08/14.0 = 3, the molecular formula is 3 times heavier than empirical; thus CH2 x 3 = C3H6 molecular formula (If you reduce this you get the empirical CH2)