Question

how many moles of calcium carbonate (CaCO3) present in 10 gram of the substance (Ca= 40 ,C= 12,O = 16)

Answer 1

$10 \div 116 = 0.086$

Answer 2

1mole of CaCO3= Gram molecular mass of CaCO3=40+12+3×16=40+12+48=100gram.=1mole

100gram of CaCO3=1mole
1gram of CaCO3=1/100 mole
10gram of CaCO3=1/100×10=0.1 mole.

Therefore, 0.1 mole of CaCO3 is present in 10grams of the substance.