Question

(a) How many moles of ammonium ions are in 0.552 g of ammonium carbonate? mol

(b) What is the mass, in grams, of 0.0279 mol of iron(III) phosphate? g

(c) What is the mass, in grams, of 4.98 1023 molecules of aspirin, C9H8O4? g

(d) What is the molar mass of a particular compound if 0.087 mol weighs 6.48 g? g/mol

Answer 1

Molar mass of (NH4)2CO3 is 96.0858 g/mol

0.552 g of ammonium carbonate = 0.552 x 2 / 96.0858 = 0.0114889 MOles of ammonium ions presents.

Molar mass of FePO4 is 150.8164 g/mol

0.0279 mol of iron(III) phosphate = 0.0279 x 150.8164 = 4.2077 gm iron(III) phosphate presents

4.98 1023 molecules of aspirin, C9H8O4 = 4.98 10²³ / 6.023 x 10²³ = 0.8268 moles presents

0.8268 moles of C9H8O4 (Molar mass of C9H8O4 is 180.1574 g/mol)= 0.8268 x 180.1574 = 148.9541 gm of C9H8O4 presents

molar mass of compound = 6.48 / 0.087 = 74.48 is the the molar mass