In choice C, because all masses are being measured on a consistently wrong balance, the faulti-
ness does not matter in the final answer.

.

..

g %%

gg

11 0 HO

11 0 7 7

(^100302)

• =

In choice D, the original mass of the hydrate would remain unchanged. However, the mass of
the anhydride would be higher than expected because the sample would lose less water than if
it had been a pure hydrate. This fits the scenario of

g %%

gg

10 HO

10 8

(^100202)

• =

with the error being consistent with the direction of the student’s results. Therefore, D is not the
correct answer.

In choice E, the original sample is wet. The freshman chemist weighed out 10 g of the hydrate,
but more weight is lost in the heating process than expected, making the final mass of the anhy-
dride lower than expected, say 6 g. Using the equation for % H 2 O shows

g %%

gg

10 HO

10 6

(^100402)

• =

which is higher than the theoretical value of 30% and in line with the reasoning that this could
NOT have caused the error.

11. When 100 grams of butane gas (C 4 H 10 , MW = 58.14) is burned in excess oxygen gas, the
    theoretical yield of H 2 O is:

A. 54 14 18 02..100 5##

B. 100 18 02^55814 ## ..

C. 13 2 100^41802 /# #. # 100 %

D.^558141802 ##.. 100

E. 100 5 18 02##58 14..

Answer: E

Begin with a balanced equation:

CH 4210 ++13 2/ O 2 " 4 CO 2 5 HO

Next, set up the equation in factor-label fashion:

.

gC H.

gC H

mole C H

mole C H

mole H O

mole H O

gH O

1 gH O

100

58 14

1

1

5

1

(^410) 18 02
410
410
410
2
2
2
###= 2
Gravimetrics