Question

The empirical formula of a compound is determined to be C2H3, and its molecular mass is found to be 54.10 g/mol. Determine the molecular formula of the compound, showing your work.

Answer 1

Find the molecular mass of C2H3

divide it by the molecular mass of 54.10g/mol, this will tell you how many times C2H3 can go into its molecular formula.

multiply the number you get from dividing the empirical formulas mass by the molecular formulas mass and multiply it by a number that will give you 1 or very close to 1.

Once you know the number that will make the ratio equal to 1 simply multiply all the atoms in the empirical formula by that.

In summery
1. Divide empirical formula molecular mass by molecular formula mass and find a number (I will call it x) you can multiply this ratio by to make it equal to one
2. Multiply every atom in the empirical formula by that number (x) to get your molecular formula.


Hope this is helpful. If you have any questions feel free to ask me or if you want me to check your answer

Answer 2

1. Find the total mass of the compound
C2H3 → 2(12.01g) + 3(1.00g) = 27.02g

2. Divide the mass of compound given in problem (empirical formula) by mass of compound found in step 1(1 mol)
54.1g ÷ 27.02g = 2.00

3. Multiply the empirical formula of the compound by a factor of 2
C2H3 x 2 → C4H6

Answer: C4H6

Answer 3

No.. This is not a practice exam. This is a module quiz from FLVS so using this answer will be an academic intergrity violation. Plus it has a obvious mistake in it.

Mr. Carlyle
FLVS Chemistry Instructor

This question is a violation of the OpenStudy Guidelines. The question is from an online school plus a violation by solicitation for members to unknowingly assist them cheating on an exam. 30653065chem30653065