Hence, molecular formula is same as empirical formula, viz.,Fe 20 3. Given that the molar mass of the oxide is 159.8 g mol -1(Atomic mass: Fe = 55.85, O = 16.00 amu) Calculation of Empirical Formula. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Thus, Cu that can be obtained from 159.5 g of CuS0 4 = 63.5 g How much copper can be obtained from 100 g of copper sulphate ( CuSO 4 ) ? (Atomic mass of Cu= 63.5 amu)Īnswer: 1 mole of CuS0 4 contains 1 mole (1 g atom) of Cu Molar mass of nitric acid HNO 3 = 1 + 14 + 48 = 63 gmol -1 Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL -1 and the mass percent of nitric acid in it is being 69%.Īnswer: Mass percent of 69% means that 100 g of nitric acid solution contain 69 g of nitric acid by mass. Molar mass of sodium acetate is 82.0245 g mol -1Īnswer: 0.375 M aqueous solution means that 1000 mL of the solution contain sodium acetate = 0.375 mole Calculate the mass of sodium acetate ( CH 3 COONa) required to make 500 mL of 0.375 molar aqueous solution. 16 g of dioxygen can combine only with 0.5 mole of carbon.C0 2 produced again is equal to 22 g. (iii) Here again, dioxygen is the limiting reactant. (ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant. Therefore,C0 2 produced from the combustion of 1 mole of carbon = 44 g. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.Īnswer: The balanced equation for the combustion of carbon in dioxygen/air is
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
Calculate the amount of carbon dioxide that could be produced when Determine the empirical formula of an oxide of Iron which has 69.9 % iron and 30.1 % dioxygen by mass.
MOLAR MASS OF NH3 MANUAL
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MOLAR MASS OF NH3 PDF
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