Relative formula mass and the mass percentage of an element
| English | Português |
|---|---|
| relative formula mass/ˈrelətɪv ˈfɔːmjʊlə mæs/ | relative formula mass |
| mass percentage | mass percentage |
What would explain this observation?
- Calcium hydroxide contains more than one oxygen and hydrogen atom. Reading its brackets correctly is essential before calculating either formula mass or calcium percentage.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Relative formula mass 相对式量, Mr, is the sum of the relative atomic masses of all atoms shown in a chemical formula. Relative atomic and formula masses are comparative numbers without gram units. For Ca(OH)₂, supplied Ar values Ca=40, O=16 and H=1 give Mr=40+2×(16+1)=74. The subscript outside the brackets multiplies both O and H.
- relative formula mass: The sum of relative atomic masses for the atoms shown in a formula; mass percentage 质量百分比: An element’s mass contribution divided by the compound’s total mass, multiplied by 100.
Which expression gives Mr of Ca(OH)₂ using Ca=40, O=16 and H=1?
An element’s percentage by mass is its total relative mass contribution divided by Mr, multiplied by 100. Calcium contributes 40 of the 74 units in Ca(OH)₂, giving 40/74×100=54.1% to three significant figures. Oxygen contributes 32 and hydrogen 2. These percentages describe the compound’s fixed mass proportions, not the proportion of different atoms counted equally.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- An element’s percentage by mass is its total relative mass contribution divided by Mr, multiplied by 100. Calcium contributes 40 of the 74 units in Ca(OH)₂, giving 40/74×100=54.1% to three significant figures. Oxygen contributes 32 and hydrogen 2. These percentages describe the compound’s fixed mass proportions, not the proportion of different atoms counted equally.
- List element, atom count, Ar and mass contribution in four columns. Add contributions for Mr, then select the requested element’s contribution for the numerator. In a balanced equation include coefficients: 2Mg + O₂ → 2MgO gives 2×24+32=80 on the left and 2×40=80 on the right. Use the periodic-table values supplied with a particular question rather than guessing more precise data.
Which two habits make the investigation or model in this case more defensible?
List element, atom count, Ar and mass contribution in four columns. Add contributions for Mr, then select the requested element’s contribution for the numerator. In a balanced equation include coefficients: 2Mg + O₂ → 2MgO gives 2×24+32=80 on the left and 2×40=80 on the right. Use the periodic-table values supplied with a particular question rather than guessing more precise data.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: MgCO₃ has supplied Ar values Mg=24, C=12 and O=16. Mr=24+12+3×16=84. Oxygen contributes 48, so oxygen percentage=48/84×100=57.1%. The carbon percentage is 12/84×100=14.3%; it is not one fifth just because one of five atoms is carbon.
With Ar C=12 and O=16, calculate the oxygen percentage by mass in CO₂ to three significant figures. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
With Ar C=12 and O=16, calculate the oxygen percentage by mass in CO₂ to three significant figures.
The result is 72.7 %. Known: MgCO₃ has supplied Ar values Mg=24, C=12 and O=16. Mr=24+12+3×16=84. Oxygen contributes 48, so oxygen percentage=48/84×100=57.1%. The carbon percentage is 12/84×100=14.3%; it is not one fifth just because one of five atoms is carbon.
Check the conclusion and its limits
- Do not multiply the whole formula by an internal subscript. A coefficient belongs to an equation amount, not the Mr of one formula unit. Percentages should total about 100, allowing rounding. A formula mass is not automatically a measured sample mass, and percentage by mass differs from percentage of atoms.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Mass percentage is found by counting all atoms as if their masses were equal. This claim is false: Do not multiply the whole formula by an internal subscript. A coefficient belongs to an equation amount, not the Mr of one formula unit. Percentages should total about 100, allowing rounding. A formula mass is not automatically a measured sample mass, and percentage by mass differs from percentage of atoms.
Relative formula mass and the mass percentage of an element: An element’s percentage by mass is its total relative mass contribution divided by Mr, multiplied by 100. Calcium contributes 40 of the 74 units in Ca(OH)₂, giving 40/74×100=54.1% to three significant figures. Oxygen contributes 32 and hydrogen 2. These percentages describe the compound’s fixed mass proportions, not the proportion of different atoms counted equally.
Mass percentage is found by counting all atoms as if their masses were equal.
Do not multiply the whole formula by an internal subscript. A coefficient belongs to an equation amount, not the Mr of one formula unit. Percentages should total about 100, allowing rounding. A formula mass is not automatically a measured sample mass, and percentage by mass differs from percentage of atoms.
The sum of relative atomic masses for the atoms shown in a formula: write the technical term.
relative formula mass means The sum of relative atomic masses for the atoms shown in a formula.