Question

Select all the correct answers.

A sample of an unknown compound has a percent composition of 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen. Which compounds could the sample be?

CH3CH3CH2O2
C2H5OH
C4H10O2
C4H12O2

Answer 1

`C_(2)`
`H_(5)`
`OH`

`C_(4)`
`H_(12)`
`O_(2)`

Step-by-step explanation:

From what we know about the unknown compound's percent composition:

• Carbon: 51.14%
• Hydrogen: 13.13%
• Oxygen: 34.73%

Let us assume we have exactly 100 grams of the unknown compound. Now that we have 100 grams of the unknown compound, the composition is:

• Carbon: 51.14 g
• Hydrogen: 13.13 g
• Oxygen: 34.73 g

Now, we need to convert the grams into moles. To do that, we divide their weight in grams by solar mass or their weight from the periodic table.

• Carbon: 51.14 ÷ 12.01 = 4.341
• Hydrogen: 13.13 ÷ 1.008 = 13.025
• Oxygen: 34.73 ÷ 16 = 2.170

Next, we divide the moles by the smallest amount to find the ratios.

• Carbon: 4.341 ÷ 2.170 = 2.000
• Hydrogen: 13.025 ÷ 2.170 = 6.002
• Oxygen: 2.170 ÷ 2.170 = 1

From this, we can gather that our formula looks may look something like
`C_(2)`
`H_(6)`
`O`. Looking at our choices, we can conclude that
`C_(2)`
`H_(5)`
`OH` equals
`C_(2)`
`H_(6)`
`O`. We can also conclude that
`C_(4)`
`H_(12)`
`O_(2)` is another possible formula for the compound since
`C_(4)`
`H_(12)`
`O_(2)` is just
`C_(2)`
`H_(6)`
`O`, but with each subscript multiplied by 2.

- 2022 Edmentum

Answer 2

C2H5OH

Step-by-step explanation:

To know the compound used as the sample, let us calculate the empirical formula for the compound. This is illustrated below:

Data obtained from the question include:

Carbon (C) = 52.14%

Hydrogen (H) = 13.13%

Oxygen (O) = 34.73%

Next, divide the above by their individual molar mass

C = 52.14/12 = 4.345

H = 13.13/1 = 13.13

O = 34.73/16 = 2.171

Next, divide the above by the smallest number.

C = 4.345/2.171 = 2

H = 13.13/2.171 = 6

O = 2.171/2.171 = 1

Therefore, empirical formula for the compound is C2H6O

Now we can re-arrage C2H6O to become C2H5OH.

Therefore, the compound used as the sample is C2H5OH