Question

Question 1

A 100 g sample with 40.92% C, 4.58% H, and 54.50% O by mass contains

A.      40.92 moles C

B.      1 mole C

C.       3.407 moles C

D.      0.2935 moles C

Part B

The ratio of the number of moles of C,H, and O in the last problem is

A.      1:2:3

B.      3.407:4.54:3.406

C.       3.407:1:2

D.      3:4:3

Part C

The simplest whole number ratio of C to H to O in problem #1 is:

A.      3:4:5

B.      1:1:1

C.       2:1.33:3

D.      3:4:3

Part D

Match the items on the left with those on the right.

A 100 g sample that is 40.92% carbon by mass

A 100 g sample that is 4.58% hydrogen by mass

A 100 g sample that is 54.50% hydrogen by mass

A 200 g sample that is 40.92% carbon

A. has 40.92 g C

B. has 4.58 g H

C. has 54.50 g H

D. has 81.84 g C

Part E

Match the following:

A 25.00 g sample with 40.92% by mass C

A 184.4 g sample with 40.92% C

A 1 g sample with 40.92% C

A 25.00 g sample with 35.46% C

A.      8.865 g C

B.      75.46 g C

C.       0.4092 g C

D.      10.23 g C

Part F

An empirical formula is always the same as the molecular formula.

A.      True

B.      False

Answer 1

1)

given 100 g sample

so

mass of C = 40.92

we know that

moles = mass / molar mass

so

moles of C = 40.92 / 12

moles of C = 3.41

so

the answer is

C) 3.407 moles C

2)

now

moles of H = 4.58 / 1= 4.58

moles of O = 54.50 / 16 = 3.40625

so

the ratio is

C:H:0 = 3.407 : 4.58 : 3.40625

so

the answer is

B) 3.407 : 4.54 : 3.40625

3)

the simplest ratio is 3:4:3

so

the answer is D) 3:4:3

4)

A 100 g sample that is 40.92% carbon by mass -----> 40.92 g C

A 100 g sample that is 4.58% hydrogen by mass ------> 4.58 g H

A 100 g sample that is 54.50% hydrogen by mass -----> 54.50 H

A 200 g sample that is 40.92% carbon   -------------. 81.84 g C

5)

A 25.00 g sample with 40.92% by mass C ---> 10.23 g C

A 184.4 g sample with 40.92% C -----> 75.46 g C

A 1 g sample with 40.92% C   ----> 0.4092 g C

A 25.00 g sample with 35.46% C ---> 8.865 g C

6)

False