Question

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical

formula?
The first step is to calculate the mass of each element in a 100-g sample of the compound. We use 100
g because it makes the calculations more straightforward.
What is the mass of each element in a 100.0-g sample of the compound?
mass H =
mass S =
mass 0 =

Answer 1

2.00, 32.7, 65.3

Step-by-step explanation:

For edge:)

Answer 2

empirical formula:
`H_2SO_4`

2 g H

32.7 g S

65.3 g O

Step-by-step explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02 : 1.02/1.02 : 4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is:
`H_2SO_4`.