The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6 Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g

a. Sample A
b. Sample B
c. Sample C
d. Sample D

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g;       H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4    or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g    H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 > 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g    H: 10 g

C: 84 / 12 = 7 > 7 / 7 = 1

H: 10 / 1 = 10    > 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g      H: 10 g

C: 90 / 12 = 7.5     > 7.5 / 7.5 = 1

H: 10 / 1 = 10  > 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6  > CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.