Empirical and molecular formulae from data
Введено| English | Значение |
|---|---|
| empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ | A formula showing the simplest whole-number ratio of atoms of each element in a compound. |
| molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ | A formula showing the actual numbers of atoms of each element in one molecule. |
Can a mass analysis reveal a formula?
- A compound's mass percentages do not directly give its atom ratio: oxygen atoms are heavier than hydrogen atoms.
- Supplement: the empirical formula 实验式 is the simplest whole-number ratio of atoms or ions; the molecular formula 分子式 gives the actual numbers of atoms in one molecule.
Turn masses into a mole ratio
- Divide each element's mass by its $A_r$. Then divide every result by the smallest result to obtain a ratio.
- If percentages are given, imagine 100 g of compound: the percentage numbers then become masses in grams. Do not round a ratio such as 1.5 down to 1.
A compound contains 24 g carbon and 6 g hydrogen. Use $A_r$: C = 12 and H = 1. Carbon amount $=24/12=2\ \mathrm{mol}$; hydrogen amount $=6/1=6\ \mathrm{mol}$. Divide by 2: C : H = 1 : 3. Its empirical formula is CH₃, not C₂₄H₆.
A compound contains 12 g carbon and 4 g hydrogen. Use Ar: C=12, H=1. Give its empirical formula using ordinary digits.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Amounts C:H=12/12 : 4/1 = 1:4, so CH₄.
Convert a simple fractional ratio
- A ratio 1 : 1.5 is equivalent to 2 : 3: multiply every part by 2. Use one common multiplier so all parts become whole numbers.
- For 5.6 g iron and 2.4 g oxygen, use $A_r$: Fe = 56, O = 16. The amounts 0.100 : 0.150 give 1 : 1.5, hence Fe₂O₃. Rounding 1.5 to 2 would give a different composition.
An atom ratio is 1 : 1.5. What is its simplest whole-number equivalent?
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Multiply every part by 2, keeping the same ratio.
Recover the molecular formula
- Find the empirical formula mass. Divide the given molecular $M_r$ by that mass to get the whole-number multiplier $k$. Multiply every empirical subscript by $k$.
- For empirical CH₃ and molecular $M_r=30$, empirical mass $=12+3=15$ and $k=30/15=2$. Molecular formula = C₂H₆. The empirical formula alone cannot tell you this multiplier.
The simplest ratio and the actual molecule are different information. Ionic lattices use formula units, not a separate molecular formula. A non-whole-number multiplier needs checking; do not silently round a large mismatch.
A molecular compound has empirical formula CH₂ and Mr 56. Use Ar: C=12, H=1. Give its molecular formula with ordinary digits.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Empirical formula mass=14. Factor=56/14=4, so C₄H₈.
Match each statement to its role.
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
A molecular formula requires information beyond the simplest ratio.
Use experimental mass differences
- If 2.4 g magnesium forms 4.0 g magnesium oxide, the oxygen gained is $4.0-2.4=1.6\ \mathrm{g}$. With $A_r$: Mg = 24, O = 16, the amounts are 0.100 mol of each element, giving MgO.
- This assumes a pure product and complete reaction. Product lost from the container, incomplete reaction or a wet product can change the measured ratio. Repeat heating, cooling and weighing to constant mass when the practical method requires it.
A 3.6 g magnesium sample forms 6.0 g of pure magnesium oxide. What mass of oxygen was gained, in g?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
Oxygen mass=6.0−3.6=2.4 g. The product mass is not the oxygen mass.
Check the formula against the data
- Recalculate its relative formula mass and compare the element contributions with the supplied masses or percentages.
- Show each stage: data → amounts → simplest ratio → empirical formula → molecular formula if a molecular $M_r$ is given.
Divide masses by $A_r$ before comparing. Simplify all parts together. Molecular formula = empirical formula multiplied by a factor found from $M_r$.
A compound contains 52.2% C, 13.0% H and 34.8% O by mass. Use Ar: C=12, H=1, O=16. Allow for the percentages being rounded. Give its empirical formula with ordinary digits.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
For 100 g: 52.2/12=4.35, 13.0/1=13.0, 34.8/16=2.175 mol. Dividing by 2.175 gives about 2:6:1.
3.6 g Mg forms 6.0 g pure MgO after complete reaction. Use Ar: Mg=24, O=16. Derive the formula from the masses and name one assumption needed.
Compare the reasoning and each criterion; equivalent wording is valid.