What is the mass in milligrams of
(a) 0.367 mol of HNO3?
(b) 245 mmol of MgO?
(c) 12.5 mol of NH4NO3?
(d) 4.95 mol of (NH4)2Ce(NO3)6 (548.23 g/mol)?
- What is the mass in grams of
(a) 3.20 mol of KBr?
(b) 18.9 mmol of PbO?
(c) 6.02 mol of MgSO4?
(d) 10.9 mmol of Fe(NH4)2(SO4)2 . 6H2O?
The Correct Answer and Explanation is :
To solve the given problems, we use the formula:
[
\text{Mass} = \text{Moles} \times \text{Molar Mass}.
]
First, calculate molar masses for each compound, then multiply by the given number of moles or millimoles (1 mmol = (10^{-3}) mol) to find the mass.
Mass in Milligrams:
- (a) 0.367 mol of HNO₃
Molar mass of HNO₃ = 1 (H) + 14 (N) + 16×3 (O) = 63 g/mol.
[
\text{Mass} = 0.367 \, \text{mol} \times 63 \, \text{g/mol} = 23.121 \, \text{g}.
]
Convert to mg: (23.121 \, \text{g} \times 1000 = 23,121 \, \text{mg}.) - (b) 245 mmol of MgO
Molar mass of MgO = 24.3 (Mg) + 16 (O) = 40.3 g/mol.
[
\text{Mass} = 245 \times 10^{-3} \, \text{mol} \times 40.3 \, \text{g/mol} = 9.8715 \, \text{g}.
]
Convert to mg: (9.8715 \, \text{g} \times 1000 = 9,871.5 \, \text{mg}.) - (c) 12.5 mol of NH₄NO₃
Molar mass of NH₄NO₃ = 14 (N) × 2 + 1×4 (H) + 16×3 (O) = 80 g/mol.
[
\text{Mass} = 12.5 \, \text{mol} \times 80 \, \text{g/mol} = 1,000 \, \text{g}.
]
Convert to mg: (1,000 \, \text{g} \times 1000 = 1,000,000 \, \text{mg}.) - (d) 4.95 mol of (NH₄)₂Ce(NO₃)₆
Molar mass = (548.23 \, \text{g/mol}).
[
\text{Mass} = 4.95 \, \text{mol} \times 548.23 \, \text{g/mol} = 2,712.7 \, \text{g}.
]
Convert to mg: (2,712.7 \, \text{g} \times 1000 = 2,712,700 \, \text{mg}.)
Mass in Grams:
- (a) 3.20 mol of KBr
Molar mass of KBr = 39.1 (K) + 79.9 (Br) = 119 g/mol.
[
\text{Mass} = 3.20 \, \text{mol} \times 119 \, \text{g/mol} = 380.8 \, \text{g}.
] - (b) 18.9 mmol of PbO
Molar mass of PbO = 207.2 (Pb) + 16 (O) = 223.2 g/mol.
[
\text{Mass} = 18.9 \times 10^{-3} \, \text{mol} \times 223.2 \, \text{g/mol} = 4.2185 \, \text{g}.
] - (c) 6.02 mol of MgSO₄
Molar mass of MgSO₄ = 24.3 (Mg) + 32.1 (S) + 16×4 (O) = 120.4 g/mol.
[
\text{Mass} = 6.02 \, \text{mol} \times 120.4 \, \text{g/mol} = 724.8 \, \text{g}.
] - (d) 10.9 mmol of Fe(NH₄)₂(SO₄)₂ · 6H₂O
Molar mass = (392.14 \, \text{g/mol}).
[
\text{Mass} = 10.9 \times 10^{-3} \, \text{mol} \times 392.14 \, \text{g/mol} = 4.273 \, \text{g}.
]
Explanation:
The key is understanding the relationship between moles, molar mass, and mass. Molar mass gives the weight of one mole of a compound, allowing direct conversion from moles to grams (or milligrams for smaller quantities). Ensure units are consistent—convert mmol to moles and grams to mg when necessary. This calculation is essential in chemistry for preparing solutions or determining reactant/product amounts.