Atoms and Molecules Worksheet-6
Calcium = 40.0%; Carbon = 12.0%; Oxygen = 48.0%
If the law of constant proportions is true, what weights of these elements will be present in 1.5 g of another sample of calcium carbonate?
(Given : Gram atomic mass of oxygen = 16 g Gram molecular mass of oxygen (O2) = 32 g)
(Given : Gram atomic mass of oxygen = 16 g, Gram molecular mass of oxygen (O2) = 32 g)
(Given : Gram atomic mass of oxygen = 16 g, Gram molecular mass of oxygen (O2) = 32 g)
Answer:
Mass of reactants = (4.0 + 10.0) = 14.0 g
Mass of products = (11.5 + 2.5) = 14.0 g
The mass of the reactants is equal to the mass of the products formed. Thus the data is in agreement with the law of conservation of mass.
Mass of reactants = (6.3 + 15.0) = 21.3 g
Mass of products = Mass of solution + Mass of carbon dioxide released.
= 18.0 g + x g
According to law of conservation of mass,
Mass of reactants = Mass of products
21.3 g = (18.0 + x)g
or x = 21.3 – 18.0 = 3.3 g
∴ Mass of carbon dioxide released = 3.3 g
In 1.5 g of another sample of Calcium carbonate 0.6 g of calcium, 0.18 g of carbon, and 0.72 g of oxygen is present.
Mass of boron in the compound = 0.096 g
Mass of oxygen in the compound = 0.144
∴ mass of O2 in grams (m) = no. of moles × M = 0.5 × (32 g) = 16 g
Mass of oxygen (O) in grams (m) = No. of moles × M = 0.5 × (16g) = 8g
Step I : Calculation of number of gram atoms of oxygen
Step II : Calculation of mass of oxygen (O) atoms
Mass of oxygen (O) atoms = Gram atomic mass of oxygen × No. Of gram atoms of oxygen
= (16g) × 0.5 = 8g
(1) 28 g N2 = 1 mole = 6.022 × 1023 molecules
14 g N2 = 0.5 mole = 6.022 × 1023 × 0.5 = 3.011 × 1023 molecules
(2) 32 g of O2 = 1 mole = 6.022 × 1023 molecules
(3) 2 g of H2 = 1mole = 6.022 × 1023 molecules
∴ 0.5 mole of NH3 will contain
3.012 × 1023 atoms of nitrogen and 9.035 × 1023 atoms of hydrogen
1.20 × 1024 atoms of sodium
45.82 g