Show:
Empirical Formulas
Molecular Formulas
Ions
Properties of Matter
Theoretical and Percent Yield
100

The empirical formula for methane 

(hint: the same as the molecular formula)

CH4

100

The molecular formula for tetranitrogen pentoxide 

N4O5

100

The charge on a ferrous ion

+2

ferric ion is +3

100

An easily separable and/or differentiable solution of 2 or more substances

Heterogenous Solution

100

The formula for percent yield

(actual/theoretical) x 100

200

The empirical formula for the following compound:

C7H14O21

CH2O3

200

Formula for magnesium chloride

MgCl2

200

The charge on a cupric ion

+2

Cuprous would be +1

200

An example of this type of property is reactivity

chemical property

200

Percent yield given the actual yield was 63.4 grams and the theoretical yield was 79.8 grams

79.5% yield

(63.4/79.8)*100%

300

The empirical formula of the following compound:

C56H72O32


C7H9O4

300

Formula for lead (IV) sulfide

PbS2

300

Charge on the sulfate ion

-2

300

A property of matter that is dependent of amount

extensive property

e.g. volume, mass, density

300

Actual yield given a percent yield of 64% and a theoretical yield of 214.8 grams

137.5 grams

(214.8)*(64/100)

400

The molecular formula for a hydrocarbon given the formula weight is 129.12 amu and the empirical formula is C2H3O

C6H9O3

6 x C = 12.01 x 6 = 72.06

9 x H = 1.01 x 9 = 9.07

3 x O = 16.00 x 3 = 48.00

72.06+9.07+48

400

Formula for cuprous nitride

Cu3N

400
Charge on the ammonium ion

+1

400

a property of matter that is independent of amount

intensive

e.g. color, smell...

400

Theoretical yield given the actual yield was 12 grams and the percent yield was 41%

29.3 grams

(12/theoretical yield) = 0.41

500

The empirical formula of glucose

Molecular: C6H12O6

Empirical: CH2O

500

The formula for ammonium bicarbonate

NH4HCO3

OR

CH5NO3

(top is more correct, chemical formula)

500

Dichromate ion formula and charge

Cr2O7 (2-)

500

The atmosphere is an example of this kind of mixture

homogeneous mixture

500

2H2 (g) + O2 (g) -> 2H2O (l)

The percent yield given I reacted 17 grams of hydrogen gas with excess oxygen and I actually produced 135 grams of H2O

89%

Theoretical yield:

(17 grams H2) * (1 mol H2/2.02 grams H2) * (2 mol H2O/2 mol H2) * (18.02 grams H2O/1 mol H2O) = 151.65 grams

135 grams (actual) / 151.65 grams (theoretical) = 0.89 = 89% yield