A compound weighing 42 grams was found to contain 12 grams of magnesium,6 grams of carbon the rest oxgen.Determine the empirical formular.i got MgCO3.am i right or wrong
3 answers
Yes, you are correct.
Mg = 24 g/mol
C = 12 g/mol
O = 16 g/mol
12/24 = 1/2 mol Mg
6/12 = 1/2 mol C
42 - 18 = 24 g O
24 g * 1 mol / 16 g = 3/2 mol O
yes, 1 1 3
MgCO3
C = 12 g/mol
O = 16 g/mol
12/24 = 1/2 mol Mg
6/12 = 1/2 mol C
42 - 18 = 24 g O
24 g * 1 mol / 16 g = 3/2 mol O
yes, 1 1 3
MgCO3
thanku so much