Empirical & Molecular Formulae — Practice Questions

Free NEET Chemistry multiple-choice questions on Empirical & Molecular Formulae. Attempt each question and reveal the answer with a full explanation.

An organic compound contains 92.3 % Carbon and 7.7 % Hydrogen. Its empirical formula is: CH CH 2 C 2H 2 CH 4 An organic compound contains 40 % carbon, 6.67 % hydrogen, and the rest is oxygen. What is the empirical formula of the compound? CH 2O CHO C 2H 2O CH 4O A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g . What is its molecular formula? C 2H 4Cl 2 CH 2Cl CHCl 3 C 2H 2Cl 4 A compound with empirical formula CH 2 has a vapour density of 42 . Its molecular formula is: C 6H 12 C 3H 6 C 2H 4 C 4H 8 A compound contains 50 % of element X (Atomic mass = 10 ) and 50 % of element Y (Atomic mass = 20 ). Its simplest formula is: X 2Y XY XY 2 X 2Y 3 An oxide of a metal contains 60.038462 percent of the metal. If the atomic mass of the metal is 15.999 3 , find the simplest formula of the oxide. M 2O 3 MO M 2O MO 2 Two elements X (at. mass 75 ) and Y (at. mass 16 ) combine to give a compound having 75.8 % of X . The formula of the compound is: X 2Y 3 XY X 2Y XY 2 An oxide of iron contains 70 % iron and 30 % oxygen by mass. Its empirical formula is: (At. mass Fe=56, O=16 ) Fe 2O 3 FeO Fe 3O 4 Fe 2O Calculate the percentage of water of crystallization in Epsom salt ( MgSO 4 7H 2O ). (At. mass Mg=24, S=32, O=16, H=1 ) 51.2 % 44.4 % 38.5 % 62.3 % Calculate the mass percent of Iron ( Fe ) in Ferrous Sulfate ( FeSO 4 ). (At. mass Fe=56, S=32, O=16 ) 36.84 % 42.12 % 25.50 % 56.00 % In a compound M 2O 3 , the metal M has an atomic mass of 27 . The percentage of oxygen in the compound is: 47.1 % 52.9 % 30 % 16 % The percentage of C, H and N in an organic compound are 40 %, 13.33 % and 46.67 % respectively. The empirical formula is: CH 4N CH 2N CH 3N C 2H 5N The molar mass of a substance is 180 g/mol . Its empirical formula is CH 2O . Its molecular formula is: C 6H 12 O 6 C 3H 6O 3 C 2H 4O 2 C 5H 10 O 5 An element A (at. wt. = 12 ) and B (at. wt. = 16 ) combine to form a compound containing 42.9 % A and 57.1 % B . The formula of the compound is: AB A 2B AB 2 A 2B 3 A hydrocarbon contains 85.7% C. If 42 mg of the compound contains 3.01 10 20 molecules, the molecular formula of the compound will be C2H4 C3H6 C6H12 C12H24 A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is : C 4A 3 C 2A 3 C 3A 2 C 3A 4 A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A xB y , then the value of x + y is in option 5 4 3 2 A gaseous hydrocarbon gives 0.66 g of CO 2 and 0.36 g of H 2O on complete combustion. The empirical formula of the hydrocarbon is: CH 2.67 which simplifies to CH 3 using integer ratios CH 2 C 2H 2 CH 4 A compound of Magnesium contains 21.9 % Magnesium, 27.8 % Phosphorus and 50.3 % Oxygen. Its empirical formula is: (At. mass: Mg=24, P=31, O=16 ) Mg2P2O7 MgP2O6 Mg3(PO4)2 MgPO3 A compound A 2B 3 contains 60 % element A by mass. The atomic mass of B is 32 u . What is the atomic mass of A ? 72 u 48 u 64 u 36 u The percentage of nitrogen in urea, NH 2CONH 2 , is approximately: 46.67 % 38.5 % 20.0 % 70.0 % What is the mass percent of Carbon in Carbon Dioxide ( CO 2 )? 27.27 % 44 % 12 % 72.73 % A compound made of two elements A and B contains 25 % of A (at. mass 12.5 ) and 75 % of B (at. mass 37.5 ). The empirical formula of the compound is: AB A 2B AB 2 A 3B A compound has an empirical formula CH 2O and a molar mass of 180 g/mol . Its molecular formula is: C 6H 12 O 6 C 3H 6O 3 C 2H 4O 2 C 5H 10 O 5 What is the percentage of Carbon in methane ( CH 4 )? 75 % 25 % 80 % 50 % An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is : [Atomic wt. of C is 12, H is 1] CH2 CH3 CH4 CH A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A = 64; B = 40; C = 32 u) A2BC2 ABC3 AB2C2 ABC4