Question

I'm confused on how to set up the problem for each question..

1.) Calculate the empirical formula of a compound if a 6.21 g sample contains 1.67g of cerium and the remainder iodine

2.) An unknown compound contains 85.64% carbon with the remainder hydrogen. It has a molar mass of 42.08 g/mol. ... I don't know how to find its empirical and molecular formulas?

3.) The molar mass of benzene, an important industrial solvent, is 78.0 g/mol and its empirical formula is CH. What is the molecular formula for benzene?

Answer 1

see the second post you did.

Answer 2

First one ,
1- percentage of cerium (Z=58) 1.67/6.21 * 100% = 26.89%
2- percentage of iodine is 100- 26.89= 73.10%
Divide each of the percentage by its mass then we get
Cerium- 26.89/140=0.192
Iodine - 73.10/126.9=0.576
Out of those two numbers divide both by the lowest no. Since 0.192 is lowest
Cerium - 0.192/0.192= 1
Iodine - 0.576/0.192= 3
Therefore empirical formula is CeI3

Answer 3

Second one same way as first you will get
C-1
H-2
Empirical formula is CH2
Empirical formula mass is 14u
Molar mass is 42.08u
Divide molar mass by empirical formula mass and you'll get n value. N=3
Multiply N to empirical formula to get molecular formula which is C3H6 (propene)