# What is C2H33O28? (Empirical Formula to Molecular Formula) [closed]

I'm doing the final question on my homework, and got an issue with this question:

"A 76.8g sample of a liquid used as a coolant in automobiles is found to contain 3.7 g of carbon, 5.1 g of hydrogen and the rest is oxygen. The formula weight of the compound is 60.0 g/mol. What is its molecular formula?"

After scouring the web, youtube, and my textbook, I can't find any reason my answer doesn't make sense. My math is as follows:

Convert each of the given masses to moles, resulting in 0.30805mol of Carbon, 5.1mol of Hydrogen, & 4.25mol of Oxygen. Dividing by the small to convert to a ratio, the result is 1mol of Carbon, 16.5mol Hydrogen, and 14mol Oxygen. Finally, to get a whole number for the ratio of moles, multiply by 2, resulting in C2H33O28.

The issue with this, is that both the mass is way too large for what we are learning, and when converting this formula to a molecular formula by dividing the given mass of the molecular formula by the mass of the empirical formula results in a 0.12 multiplier. Multiplying each of the subscripts by 0.12 gives an odd result of C0.237H4O3.

I have no idea what I did wrong here besides maybe rounding, but after trying this four separate times with different rounding. If anyone can provide any insight whatsoever, I would be ecstatic, thank you so much!

• Are you sure there isn't a typo somewhere in the question? Your approach seems reasonable, but the resulting formula is not a sensible molecule. – Tyberius Jan 22 at 2:37
• @Rocky The data is wrong. It seems the sample mass is off. Don't bother to waste your time and ask your teacher about the numbers. – M. Farooq Jan 22 at 2:44
• Alright, I'll ask her tomorrow, thank you for the help! – Rocky Jan 22 at 2:46
• Try "30.7 g of carbon". – Andrew Jan 22 at 14:12
• Notice there is said liquid, not compound. In the car cooling context, it looks like water solution of ethylenglykol. From element ratio should be possible to calculate concentration. Unfortunately, M=62 g/mol. – Poutnik Jan 22 at 20:29