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The mass percentage of a given element in a compound must always be (greater/less) than \(100 \%\).

Short Answer

Expert verified
The mass percentage of a given element in a compound must always be less than or equal to \(100\%\) because it is calculated using the formula: Mass Percentage = \(\frac{\text{Mass of the element}}{\text{Total mass of the compound}}\) × 100%, and the mass of the element can never be greater than the total mass of the compound.

Step by step solution

01

Understanding Mass Percentage

Mass percentage refers to the percentage of a certain element's mass present in a compound relative to the total mass of the compound.
02

Calculating Mass Percentage

The formula for mass percentage is given by: Mass Percentage = \(\frac{\text{Mass of the element}}{\text{Total mass of the compound}}\) × 100%
03

Analyzing the Formula

As the formula shows, the mass of the element must be less than or equal to the total mass of the compound. It can never be more than the total mass of the compound. As the element's mass is divided by the total mass, the result of this division will always be less than or equal to 1.
04

Conclusion

When we multiply the result of the division (which is 1 or less) by 100%, we find the mass percentage. Therefore, the mass percentage of a given element in a compound must always be less than or equal to 100%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculating Mass Percentage
Understanding how to calculate the mass percentage of an element in a compound is crucial for various applications in chemistry. In simple terms, mass percentage is just what it sounds like – it represents the proportion of the total mass of a compound that is made up by a particular element.

For instance, to calculate the mass percentage of oxygen in water (H2O), you'd first determine the molar mass of water, which is approximately 18 grams per mole (16 grams for oxygen and 1 gram for each hydrogen atom). Given that oxygen's part is 16 grams, you'd use the formula provided in the textbook solution:
\[ \text{Mass Percentage} = \left( \frac{\text{Mass of the element}}{\text{Total mass of the compound}} \right) \times 100\% \]
This formula essentially tells us how to create a ratio of the element's mass to the total mass, and then convert it into a percentage by multiplying by 100. It's a handy way to express the composition that easily translates into a visual understanding of a compound's make-up.
Composition of Compounds
The composition of compounds is a fundamental concept that reveals the specific ratios of elements that make up a chemical substance. Each compound has a unique composition showcased by its chemical formula. For example, carbon dioxide (CO2) consists of one carbon atom and two oxygen atoms.

When considering mass, the composition tells us how much of each element's mass contributes to the total mass of the compound. This information can be particularly valuable when converting between the mass of a compound and the amount of an element it contains. In laboratories and industrial processes, knowing the composition of compounds is essential for predicting product yields, understanding reactions, and ensuring safety.
Stoichiometry
Stoichiometry is a branch of chemistry that involves the calculation of reactants and products in chemical reactions. It is a quantitative reflection of the principle that matter is conserved in every chemical reaction. This discipline is based on the fact that, due to the conservation of mass, the mass of the products in a chemical reaction equals the mass of the reactants.

Stoichiometry allows chemists to predict the amount of substances consumed or produced in a given reaction using the balanced equation. It incorporates the concept of the mole (a standard unit for amount of substance in chemistry) and molar ratios from the balanced equation to perform calculations. By mastering stoichiometry, students are able to tackle a wide range of problems, from calculating yields to understanding limiting reagents, which dictate the extent to which a reaction will go forward.

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Most popular questions from this chapter

Calculate the mass in grams of hydrogen present in 2.500 g of each of the following compounds. a. benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) b. calcium hydride, \(\mathrm{CaH}_{2}\) c. ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) d. serine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{3} \mathrm{N}\)

A compound has the following percentages by mass: aluminum, 32.13\%; fluorine, 67.87\%. Calculate the empirical formula of the compound.

Calculate the mass in grams of each of the following samples. a. 2.41 millimol of potassium nitrate (1 millimol \(=\) \(1 / 1000 \mathrm{mol})\) b. 8.91 mol of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. 0.0141 mol of calcium oxide d. \(1.91 \mathrm{mol}\) of gold(III) bromide e. 0.0000117 mol of water f. 2.68 mol of silver nitrate

Calculate the percent by mass of the element mentioned first in the formulas for each of the following compounds. a. sodium azide, \(\mathrm{NaN}_{3}\) b. copper(II) sulfate, \(\mathrm{CuSO}_{4}\) c. gold(III) chloride, \(\mathrm{AuCl}_{3}\) d. silver nitrate, \(\mathrm{AgNO}_{3}\) e. rubidium sulfate, \(\mathrm{Rb}_{2} \mathrm{SO}_{4}\) f. sodium chlorate, \(\mathrm{NaClO}_{3}\) g. nitrogen triiodide, \(\mathrm{NI}_{3}\) h. cesium bromide, \(\operatorname{CsBr}\)

Calculate the molar mass for each of the following substances. a. adipic acid, \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) b. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{N}_{4} \mathrm{O}_{2}\) c. eicosane, \(\mathrm{C}_{20} \mathrm{H}_{42}\) d. cyclohexanol, \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) e. vinyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) f. dextrose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

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