140k views
4 votes
Can someone please help me with the empirical formula of

C:63.15% H:5.30%
O:31.55%

I am getting 3C 3H 1O but that’s not one of the options on my sheet so I don’t know if I’m doing it wrong because all my other ones are done right

2 Answers

4 votes

Answer:

C12H12NO3

Step-by-step explanation:

User Brian Petro
by
5.2k points
2 votes

Answer:

The answer is C8H8O3.

You rounded in a place where you shouldn't have (see explanation below)!

Step-by-step explanation:

These are percents by mass. Pretend you have a 100g sample. That means you have:

63.15g C

5.30g H

31.55g O

Divide each by molar mass to find the moles of each:

C: 63.15 / 12 = 5.26 mol

H: 5.30 / 1 = 5.30 mol

O: 31.55 / 16 = 1.97 mol

Now divide each of these values by the lowest of the three to find the ratio between each element in the compound:

C: 5.26 / 1.97 ≈ 2.67

H: 5.30 / 1.97 ≈ 2.69

O: 1.97 / 1.97 = 1

These aren't whole numbers and you can't just round up. If the numbers you ended with were something around 2.98, I would be fine with rounding to 3. But here 2.67 is too different from 3 to the point where we're actually considering a completely different compound.

So, we need to multiply them all by some factor to get whole numbers that will make sense in a chemical formula. If you express 2.67 and 2.69 as fractions, you get about 8/3, and 1 can be represented as 3/3, so 3 is the LCD of all three numbers. Multiply each of the numbers by 3 to get your final answer:

C: 2.67 x 3 ≈ 8

H: 2.69 x 3 ≈ 8

O: 1 x 3 = 3

Side note: if you take the percent masses of the empirical formula you came up with, you'll see they're quite different from the one the question provided, which is why you can't round in this situation:

Finding mass using your empirical formula:

C: 3 x 12 = 36

H: 3 x 1 = 3

O: 1 x 16 = 16

Total mass = 36 + 3 + 16 = 55

% mass:

C: 36/55 = 65.45% (vs. 63.15%)

H: 3/55 = 5.45% (vs. 5.30%)

O: 16/55 = 29.09% (vs. 31.55%)

The values differ by more than 2% in some cases!

User BeUndead
by
5.5k points