Question

What is the mass percent of carbon in propanoic acid `(C_2H_5COOH)`?

The molar mass of `C = 12.01, H = 1.0079`, and `O=16.00`.

Answer 1

`%"(C in C"_2"H"_5"COOH)"= 48.64%`

Explanation:

`"A"_r "(C)"= 12.01`
`"M"_r "(C"_2"H"_5"COOH)"= 3xx12.01 + 6xx1.0079 + 2xx16.00 = 74.0774`

`%"(C in C"_2"H"_5"COOH)"= ("A"_r"(C)"xx"N"°"(C in C"_2"H"_5"COOH)")/("M"_r"(C"_2"H"_5"COOH)")xx100 = (12.01xx3)/74.0774xx100= 48.64%`