3) What is the % by mass of oxygen in H2SO4 ? [ GFM=98]

ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS

EQUALS MASS

SUBTOTAL

H 2 X 1.00794 g/mol

= 2.01588 g

S 1 X 32.01

g/mol

= 32.01 g

O 4 X 15.99

g/mol

= 63.9976 g

% COMP (MASS) = PART X 100 WHOLE

% Oxygen = 63.9976 X 100 98

= 65.3 % ~ 65 %

#5) A hydrate is a compound that includes water molecules within its crystal structure. During an experiment to determine the percent by mass of water in a hydrated crystal, a student found the mass of the hydrated crystal to be 4.10 grams. After heating to constant mass, the mass was 3.70 grams. What is the percent by mass of water in this crystal?

100Mass water

=Mass hydratex

This is a modified version of part/whole x 100 from the reference tables!

% H2O =

1004.10 g – 3.70 g

=4.10 gx% H2O = 9.8 %

water

7) Given the reaction:

6 CO2 + 6 H2O 1 C6H12O6 + 6 O2

a) What is the total number of moles of water needed to make 2.5 moles of C6H12O6? H2O . =

C6H12O6 . .

62.5x

1 X = 15 moles of water

Theoretical mole ratio

#13) A compound contains 0.5 moles of sodium, 0.5 moles of nitrogen, and 1.0 moles of hydrogen. The empirical formula of the compound is

Na0.5N0.5H1.0

Subscripts can not be decimals, to get rid of a 0.5 decimal

multiply all subscripts by 2!

Na0.5x2N0.5x2H1.0x2

The new subscripts are:

Multiply subscripts by 2

Na1N1H2

NaNH2

or

#15) A compound contains 40% CALCIUM, 12% CARBON and 48% OXYGEN by mass. What is the empirical formula of this compound?

ELEMENT

MASS / ATOMIC MASS RAW RATIO

DIVIDE BY SMALLEST

SUBSCRIPTRATIO

Ca 40 g / 40.08g/mol =0.998 / 0.998 = 1.0

C 12 g / 12.01g/mol =0.999 / 0.998 = 1.0

O 48 g / 15.99g/mol =3.00 / 0.998 = 3.0

Assume 100g of the sample, this will allow you to assume 40% is 40 grams. Total mass does NOT affect % composition.

Ca1C1O3

CaCO3