纯净物的元素组成

AP 化学· AP 化学 CED — 原子结构与性质· 14 分钟阅读

1. 质量分数组成★★☆☆☆⏱ 3 min

📘 定义

质量分数组成

纯净化合物总质量中，某一特定组成元素所占的百分比。根据定比定律，所有纯净物中每种元素的质量分数都是固定的。

质量分数可以直观地告诉你100g化合物样品中所含该元素的质量，这会简化后续实验式的计算，公式为：

\text{Mass percent of X} = \frac{(\text{subscript of X} \times \text{Average atomic mass of X})}{\text{Molar mass of the compound}} \times 100\%

📐 例题

碳酸钙 $CaCO_{3}$ 中氧元素的质量分数是多少？结果保留一位小数？

1
从元素周期表查找平均原子质量：
2
$Ca = 40.08 \text{ g/mol}, C = 12.01 \text{ g/mol}, O = 16.00 \text{ g/mol}$
3
计算1 mol $CaCO_{3}$ 中氧元素的总质量（氧的下标为3）：
4
$3 \times 16.00 = 48.00 \text{ g/mol}$
5
计算 $CaCO_{3}$ 的总摩尔质量：
6
$40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}$
7
代入质量分数公式：
8
$\frac{48.00}{100.09} \times 100\% = 47.9\%$
9
碳酸钙中氧元素的质量分数为47.9%。

Exam tip:

计算结束后，务必将化合物中所有元素的质量分数相加；如果总和与100%的差值超过0.5%，说明存在计算错误。

2. 由质量数据推导实验式★★★☆☆⏱ 4 min

📘 定义

实验式

化合物中各元素原子的最简整数比。离子化合物的化学式通常就是实验式，而分子化合物的分子式是实验式的整数倍。

要根据实验数据计算实验式，请遵循以下四个核心步骤：

Convert mass of each element to moles using $n = \frac{m}{M}$
Divide all mole values by the smallest mole value to get a preliminary ratio
Multiply all ratios by a whole number to convert any fractional ratios to whole numbers
Use the whole numbers as subscripts for the empirical formula

📐 例题

一份7.50g的氮氧化物纯净样品中含有2.30g氮。该化合物的实验式是什么？

1
通过差值法计算氧元素的质量：
2
$7.50 \text{ g} - 2.30 \text{ g} = 5.20 \text{ g O}$
3
将质量转换为物质的量：
4
$n(\text{N}) = \frac{2.30 \text{ g}}{14.01 \text{ g/mol}} \approx 0.164 \text{ mol}; \quad n(\text{O}) = \frac{5.20 \text{ g}}{16.00 \text{ g/mol}} \approx 0.325 \text{ mol}$
5
将两个数值都除以最小的物质的量（0.164 mol）：
6
$\text{N} = \frac{0.164}{0.164} = 1.00; \quad \text{O} = \frac{0.325}{0.164} \approx 1.98 \approx 2$
7
N和O的整数比为1:2，因此实验式为：
8
$\text{NO}_2$

Exam tip:

如果比值与整数的差在0.05以内（例如1.98接近2），直接四舍五入为整数；只有当比值明显是1.25、1.33、1.5这类分数时，才需要乘以整数化为整数。

3. 由实验式和摩尔质量推导分子式★★☆☆☆⏱ 3 min

得到分子化合物的实验式后，如果知道该化合物的实验摩尔质量（AP题目中通常由质谱给出），就可以求出实际的分子式。因为分子式是实验式的整数倍，所以它的摩尔质量也是实验式摩尔质量的相同倍数。倍数 $n$ 的计算公式为：

n = \frac{\text{Molar mass of molecular compound}}{\text{Molar mass of empirical formula}}

$n$ will always be a whole number greater than or equal to 1. If $n = 1$ , the empirical and molecular formulas are identical. After finding $n$ , multiply all subscripts in the empirical formula by $n$ to get the final molecular formula.

📐 例题

A molecular compound has an empirical formula of $NO_{2}$ , and an experimental molar mass of 92.01 g/mol. What is its molecular formula?

1
Calculate the molar mass of the empirical formula $NO_{2}$ :
2
$14.01 + (2 \times 16.00) = 46.01 \text{ g/mol}$
3
Calculate the multiplier $n$ :
4
$n = \frac{92.01}{46.01} \approx 2$
5
Multiply each subscript in the empirical formula by 2: N = $1 \times 2 = 2$ , O = $2 \times 2 = 4$ . The molecular formula is:
6
$\text{N}_2\text{O}_4$

Exam tip:

Always confirm that the molar mass of your calculated molecular formula matches the given molar mass within rounding error before writing your final answer.

4. Combustion Analysis for Organic Compounds★★★★☆⏱ 4 min

Combustion analysis is an experimental technique used to determine the elemental composition of pure organic compounds (most commonly compounds made of C, H, and O). In the experiment, a known mass of the organic compound is burned completely in excess oxygen, and all $CO_{2}$ and $H_{2} O$ produced are absorbed by pre-weighed materials. All carbon in the original compound becomes $CO_{2}$ , and all hydrogen becomes $H_{2} O$ , so we can calculate the mass of C and H in the original compound from product masses. Any remaining mass of the original compound is oxygen, since excess oxygen from the reaction does not contribute to the original sample mass.

📐 例题

A 0.500 g sample of a pure organic compound containing only C, H, and O produces 0.733 g $CO_{2}$ and 0.300 g $H_{2} O$ in combustion analysis. What is the empirical formula?

1
Calculate moles and mass of C (all C from the original sample becomes $CO_{2}$ ):
2
$n(\text{C}) = n(\text{CO}_2) = \frac{0.733 \text{ g}}{44.01 \text{ g/mol}} \approx 0.01666 \text{ mol}; \quad m(\text{C}) = 0.01666 \times 12.01 \approx 0.200 \text{ g}$
3
Calculate moles and mass of H (each mole of $H_{2} O$ has 2 moles of H from the original compound):
4
$n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{0.300 \text{ g}}{18.015 \text{ g/mol}} \approx 0.0333 \text{ mol}; \quad m(\text{H}) = 0.0333 \times 1.008 \approx 0.0336 \text{ g}$
5
Calculate mass and moles of O by difference from the original sample mass:
6
$m(\text{O}) = 0.500 - 0.200 - 0.0336 = 0.2664 \text{ g}; \quad n(\text{O}) = \frac{0.2664}{16.00} \approx 0.01665 \text{ mol}$
7
Divide all mole values by the smallest mole value (0.01665 mol) to get a 1:2:1 ratio, so the empirical formula is:
8
$\text{CH}_2\text{O$

Exam tip:

Don’t forget to multiply the moles of $H_{2} O$ by 2 to get moles of H; forgetting this step is the most common mistake in combustion analysis problems.

5. 常见陷阱

错误做法:

When calculating empirical formula from percent composition, you use the percent values directly as moles without converting to mass first.

原因:

Students implicitly assume a 100 g sample but forget percent values are percentages, not masses, leading to incorrect mole calculations.

正确做法:

Always explicitly assume a 100 g sample, so each percent value becomes the mass of the element in grams before converting to moles.

错误做法:

In combustion analysis, you calculate mass of O by adding oxygen from $CO_{2}$ and $H_{2} O$ instead of using mass difference.

原因:

Students forget that almost all oxygen in the products comes from the excess oxygen used for combustion, not the original compound.

正确做法:

Always calculate mass of O in the original compound by subtracting mass of C and H from the total mass of the original sample.

错误做法:

You round a 1.33 mole ratio to 1 instead of multiplying all ratios by 3 to get a whole-number ratio.

原因:

Students are eager to round to whole numbers and miss common simple fractions that require scaling.

正确做法:

Recognize common fractions (0.25 = 1/4, 0.33 = 1/3, 0.5 = 1/2, 0.66 = 2/3) and multiply all ratios by the denominator of the fraction before rounding.

错误做法:

You calculate n as empirical molar mass divided by molecular molar mass, getting a value less than 1, then round incorrectly.

原因:

Students mix up the order of division in the multiplier formula.

正确做法:

Memorize the order: $n = \frac{larger molar mass (molecular)}{smaller molar mass (empirical)}$ , which always gives a whole number ≥ 1.

错误做法:

You multiply only the first subscript in the empirical formula by n when calculating molecular formula.

原因:

Students rush and forget to apply the multiplier to all elements.

正确做法:

Always multiply every subscript in the empirical formula by n, not just the first one.

错误做法:

You calculate mass percent of an element using only the atomic mass once, ignoring the subscript.

原因:

Students forget the subscript indicates how many atoms of the element are in one formula unit.

正确做法:

Always multiply the atomic mass of each element by its subscript when calculating total mass of the element.

6. 速查表

Category	Formula / Process	Notes
Mass percent of element X	$Mass % X = \frac{( subscript X \times A _{r} ( X ))}{M ( compound )} \times 100%$	Sum of all mass percents must equal ~100%
Empirical formula from mass	Convert mass to moles; 2. Divide by smallest mole; 3. Scale to whole numbers	Assume 100 g sample if given mass percent
Molecular formula multiplier	$n = \frac{M ( molecular )}{M ( empirical )}$	n is always whole number ≥ 1; multiply all subscripts by n
Moles of C (combustion)	$n (C) = n (CO_{2})$	All C from original compound becomes CO₂
Moles of H (combustion)	$n (H) = 2 \times n (H_{2} O)$	1 mole H₂O has 2 moles H from original compound
Mass of O (C/H/O combustion)	$m (O) = m_{sample} - m (C) - m (H)$	Most O in products comes from excess combustion O₂
Empirical vs Molecular Formula	N/A	Ionic compounds always use empirical formula; only molecular compounds have distinct molecular formulas

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · MCQ
由数据计算实验式
2022 · FRQ
燃烧分析问题

下一步

Elemental composition of pure substances is the foundational link between macroscopic mass measurements and microscopic atomic composition, which is required for nearly all quantitative calculations in AP Chemistry. Mastery of these calculation techniques allows you to connect experimental lab data to the chemical identity of unknown pure compounds, a core skill that appears across every unit of the AP Chemistry course. Immediately after mastering this topic, you will move on to composition of mixtures, then apply your empirical formula skills to stoichiometry of chemical reactions, where you will use these elemental ratios to calculate reactant and product yields. Without correctly determining elemental composition and empirical formulas, you will not be able to solve limiting reactant problems, titration calculations, or any other quantitative problem that relies on mole ratios of compounds.