What Information Is Needed to Calculate the Percent Composition of a Compound?
Per centum Composition of Compounds
The percent limerick (past mass) of a compound can be calculated by dividing the mass of each chemical element by the total mass of the compound.
Learning Objectives
Translate between a molecular formula of a compound and its pct composition by mass
Key Takeaways
Fundamental Points
- The atomic composition of chemical compounds can exist described in a variety of ways, including molecular formulas and percent composition.
- The percent composition of a compound is calculated with the molecular formula: dissever the mass of each element found in one mole of the compound by the total molar mass of the chemical compound.
- The pct limerick of a compound can be measured experimentally, and these values can exist used to determine the empirical formula of a compound.
Key Terms
- percent by mass: The fraction, by weight, of one element of a compound.
The atomic limerick of chemical compounds tin exist described using a diverseness of notations including molecular, empirical, and structural formulas. Another user-friendly way to describe atomic composition is to examine the pct composition of a compound by mass.
Percent Composition by Mass
Percent composition is calculated from a molecular formula by dividing the mass of a single element in ane mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.
For case, butane has a molecular formula of C4H10. Butane's percent composition tin be calculated every bit follows:
- Mass of H per mol butane: [latex]10\text{ mol H}\cdot\frac{ane.00794\text{ 1000}}{1\text{ mol H}} = 10.079\text{ m H}[/latex]
- Mass of C per mol butane: [latex]4\text{ mol C}\cdot\frac{12.011\text{ g C}}{1\text{mol C}}=48.044\text{g C}[/latex]
- Mass per centum H in butane: [latex]\frac{x.079\text{g H}}{58.123\text{ g butane}}\cdot100=17.three\%\text{ H}[/latex]
- Mass pct C in butane: [latex]\frac {48.044\text{g C}}{58.123\text{ g butane}}\cdot100=82.7\%\text{ C}[/latex]
Therefore, the atomic composition of butane can also be described as 17.iii% hydrogen and 82.7% carbon, and, as expected, these values sum to 100%.
Butane: The structural formula of butane.
In practice, this calculation is frequently reversed. Mass percents tin be determined experimentally via elemental analysis, and these values tin exist used to summate the empirical formula of unknown compounds. However, this data is bereft to decide the molecular formula without additional data on the compound's molecular weight.
Combustion Analysis
Combustion assay is commonly used to determine the relative ratios of carbon, hydrogen, and oxygen in organic compounds.
Learning Objectives
Draw the process of combustion analysis.
Key Takeaways
Cardinal Points
- Combustion is the process of burning an organic compound in oxygen to produce free energy, carbon dioxide, and h2o vapor.
- In combustion analysis, a sample of known mass is combusted, and the resulting carbon dioxide and h2o vapor are captured and weighed.
- The relative amounts of carbon, hydrogen, and oxygen in the starting compound can exist determined from the masses of the products of the combustion reaction.
- Combustion assay tin can thus be used to determine the empirical formula of an unknown organic chemical compound.
Cardinal Terms
- combustion analysis: Using burning to determine the elemental composition of an organic chemical compound. The compound is burned, the products are collected and weighed, and the composition is determined.
- combustion: A process wherein a fuel is combined with oxygen, unremarkably at loftier temperature, releasing heat, carbon dioxide and water vapor.
Combustion assay is an elemental analytical technique used on solid and liquid organic compounds. It can determine the relative amounts of carbon, hydrogen, oxygen in compounds, and occasionally can as well place the amounts nitrogen and sulfur in compounds. This technique was invented by Joseph Louis Gay-Lussac.
Combustion
Combustion assay is commonly used to analyze samples of unknown chemic formula. It requires only milligrams of a sample. The sample is weighed and then fully combusted at a high temperature in the presence of excess oxygen, which produces carbon dioxide and water.
Ane instance of a uncomplicated combustion reaction is the combustion of methane:
[latex]\text{CH}_{iv} + 2\text{O}_{2} \rightarrow\text{CO}_{2} + 2\text{H}_{two}\text{O} + \text{free energy}[/latex]
Combustion: Energy is released in the form of flames as the fuel combusts.
Another mutual example of combustion is the burning of wood to produce thermal free energy. When 1 mole of propane (C3Height) is burned in excess oxygen, 3 moles of CO2 and 4 moles of HiiO are produced.
Combustion Analysis
In combustion analysis, the products, carbon dioxide and water vapor, are trapped past absorption onto reactive solids located in tubes in a higher place the reaction vessel. These tubes tin can then be weighed to decide the captivated masses of carbon dioxide and h2o.
- The mass of carbon in the starting textile is determined past a 1:1 ratio with the mass of carbon dioxide produced (as in the combustion reaction for methane already displayed).
- The initial hydrogen mass is determined by a ii:1 ratio with the amount of water produced.
The data and the ratios tin can then be used to calculate the empirical formula of the unknown sample. Combustion analysis tin can too be performed using a CHN analyzer, which uses gas chromatography to analyze the combustion products.
Source: https://courses.lumenlearning.com/boundless-chemistry/chapter/compound-composition/
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