Question

A compound is 2% H, 32.7% S, and 65.3% O by mass. What is the subscript on the O in the empirical formula for this compound?

Answer 1

2/1.01=1.980=2
32.7/32.07=1.020=1
65.3/16.00=3.969=4

H2SO4. oxygen is 1

Answer 2

The subscrip for O is 4

Step-by-step explanation:

The first step for this question is to assume that we have 100 g of the unknown compound. If we have 100g then we will have:

2 g of H

32.7 g of S

65.3 g of O

The next step is find the moles for each atom. So, we have to use the molar mass of each atom.

1 g/mol of H

32 g/mol for S

16 g/mol for O

Now, we can find the moles if we divide by the molar mass:

`2~g~H(1~molH)/(1~gH)=2~molH`

`32.7~g~H(1~molS)/(32~gS)=1.02~molS`

`65.3~g~H(1~molO)/(16~gO)=4.08~molO`

Then we have to divide the smallest mol number, so:

`(2~molH)/(1.02)=2~molH`

`(1.02~molS)/(1.02)=1~molS`

`(4.08~molO)/(1.02)=4~molO`

So, the formula would be:
`H_2SO_4` with a "4" for oxygen.