U1.3 · Elemental Composition of Pure SubstancesPractice · 纯净物的元素组成
Lessons covered:U1.3 Elemental Composition of Pure Substances
- 1Multiple choice
A sample of a hydrocarbon contains 2.40 g of C and 0.600 g of H. Which compound has the same empirical formula?
Explanationn(C) = 2.40 ÷ 12.01 = 0.200 mol and n(H) = 0.600 ÷ 1.008 = 0.595 mol, so C : H = 1 : 2.98 ≈ 1 : 3, empirical formula . reduces to ; , (), and don’t.
- 2Multiple choice
Which of these compounds has the greatest mass percent of nitrogen?
ExplanationMass percent = total mass of N in the formula ÷ formula mass. 82.2%, 87.4%, 35.0%, 46.7%. has only light H atoms besides N, so its nitrogen percent is highest; O and C dilute the others.
- 3Numeric
What is the mass percent of water in Epsom salt, ?
%ExplanationFormula mass = 120.37 + 7 × 18.02 = 246.48 g/mol, of which water is 7 × 18.02 = 126.14 g. Mass percent = 126.14 ÷ 246.48 × 100% = 51.2%: more than half the mass is water.
- 4Multiple choice
Four pure samples containing only carbon and oxygen were analyzed, with the results shown. Which two samples could be the same compound?
Sample C (g) O (g) A 3.00 4.00 B 1.20 3.20 C 0.60 0.80 D 2.40 4.80 ExplanationBy the law of definite proportions, a pure compound always has the same mass ratio of its elements, whatever the sample size. Mass ratio of O to C: A 1.33, B 2.67, C 1.33, D 2.00. A and C have the same ratio (both are CO); B () and D differ.
- 5Multiple choice
A compound is 54.5% C, 9.1% H, and 36.4% O by mass, and its molar mass is about 88 g/mol. What is its molecular formula?
ExplanationTake 100 g: n(C) = 4.54, n(H) = 9.03, n(O) = 2.28 mol. Divide by the smallest, 2.28: about 2 : 4 : 1, empirical formula (44.05 g/mol). 88 ÷ 44.05 ≈ 2, so the molecular formula is . is only the empirical formula.
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