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Chapter 8: Problem 44

If the amount of a sample doubles, what happens to the percent composition of each element in the compound?

Short Answer

Expert verified
The percent composition of each element in a compound remains unchanged when the amount of the sample is doubled. In this case, the original and new percent compositions of H and O in water are H = 11.11% and O = 88.89%, which stay the same when the amount of water is doubled.

Step by step solution

01

Find the molar mass of the compound

Calculate the molar mass of the compound (in this case, water) by adding the atomic weights of its constituent elements: Molar mass of H鈧侽 = 2 * (Atomic weight of H) + Atomic weight of O Molar mass of H鈧侽 = 2 * 1 + 16 = 18 g/mol
02

Calculate the percentage composition of each element

Calculate the percentage composition of each element in the compound using the formula: Percent composition of an element = (Atomic weight of the element / Molar mass of the compound) * 100 Percent composition of Hydrogen in H鈧侽 = (2 * 1 / 18) * 100 = (2 / 18) * 100 = 11.11% Percent composition of Oxygen in H鈧侽 = (16 / 18) * 100 = 88.89% Original percent composition: H = 11.11%, O = 88.89%
03

Double the amount of the sample

Now let's double the amount of water and observe whether the percentage composition changes. Doubling the amount of water means considering two molecules of water (2H鈧侽) instead of one. Molar mass of 2H鈧侽 = 2 * (Molar mass of H鈧侽) = 2 * 18 = 36 g/mol
04

Calculate the new percentage composition

Calculate the new percentage composition of each element in the doubled compound: Percent composition of Hydrogen in 2H鈧侽 = (4 * 1 / 36) * 100 = (4 / 36) * 100 = 11.11% Percent composition of Oxygen in 2H鈧侽 = (2 * 16 / 36) * 100 = (32 / 36) * 100 = 88.89% New percent composition: H = 11.11%, O = 88.89%
05

Compare the percentage compositions

Now let's compare the original percent composition (before doubling the amount) with the new percent composition (after doubling the amount): Original percent composition: H = 11.11%, O = 88.89% New percent composition: H = 11.11%, O = 88.89% We can observe that the percentage composition of elements in the compound remains unchanged when the amount of the sample is doubled.

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Key Concepts

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

Molar Mass Calculation
Understanding the molar mass of a compound is a fundamental concept in chemistry that plays a vital role in various calculations. Molar mass, often expressed in grams per mole (g/mol), is the mass of one mole of a substance. It is calculated by summing the atomic weights of all the atoms in a molecule of the compound.

For example, water (H鈧侽) has a molar mass that is the sum of twice the atomic weight of Hydrogen (H) and the atomic weight of Oxygen (O). Using the periodic table, the atomic weight of H is approximately 1, and the atomic weight of O is about 16. The calculation for the molar mass of water is as follows:

Molar mass of H鈧侽 = 2 * (Atomic weight of H) + (Atomic weight of O) = 2 * 1 + 16 = 18 g/mol.

This calculation involves basic arithmetic but is essential for understanding the composition and reactions of the compound.
Atomic Weight
The atomic weight, or atomic mass, represents the average mass of atoms of an element, measured in atomic mass units (amu). It accounts for the relative abundance of various isotopes of an element in a naturally occurring sample. The atomic weight values for each element are found on the periodic table and are crucial for many chemistry calculations, including the determination of molar mass.

Take Hydrogen as an example, its atomic weight is about 1 amu. This value is an average that takes into consideration the natural abundance of Hydrogen's isotopes, such as protium (鹿H), deuterium (虏H), and tritium (鲁H), though protium is the most abundant. When calculating percent compositions or preparing solutions, the atomic weight serves as the 'building block' for the calculation.
Chemical Compound Analysis
Chemical compound analysis involves determining the percent composition of each element in a compound, among other analytical techniques. Percent composition is calculated by dividing the total atomic weight of each element in a molecule by the molar mass of the compound and then multiplying the result by 100 to convert it to a percentage.

This is not only foundational for understanding the compound's molecular makeup but is also utilized in stoichiometry during chemical reactions, in nutrition to determine the percentage of elements in food, and in material science.

Example Calculation

For water:Percent composition of Hydrogen = (Atomic weight of H * Number of H atoms / Molar mass of H鈧侽) * 100
For Oxygen:Percent composition of Oxygen = (Atomic weight of O / Molar mass of H鈧侽) * 100

As illustrated in the exercise, this percentage composition remains constant regardless of the quantity of the substance at hand. Therefore, the proportion of each element in a chemical compound does not change with the size of the sample, which is a cornerstone principle in chemistry called the law of definite proportions.

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Most popular questions from this chapter

A compound was analyzed and was found to contain the following percentages of the elements by mass: nitrogen, \(11.64 \% ;\) chlorine, \(88.36 \%\). Determine the empirical formula of the compound.

Although exact isotopic masses are known with great precision for most elements, we use the average mass of an element's atoms in most chemical calculations. Explain.

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds. a. adipic acid, \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) b. ammonium nitrate, \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) c. caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) d. chlorine dioxide, \(\mathrm{ClO}_{2}\) e. cyclohexanol, \(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{OH}\) f. dextrose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) g. eicosane, \(\mathrm{C}_{20} \mathrm{H}_{42}\) h. ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Calculate the mass in grams of each of the following samples. a. 10,000,000,000 nitrogen molecules b. \(2.49 \times 10^{20}\) carbon dioxide molecules c. 7.0983 moles of sodium chloride d. \(9.012 \times 10^{-6}\) moles of 1,2 -dichloroethane, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\)

Find the item in column 2 that best explains or completes the statement or question in column 1 Column 1 (1) \(1 \mathrm{amu}\) (2) 1008 amu (3) mass of the "average" atom of an element (4) number of carbon atoms in \(12.01 \mathrm{~g}\) of carbon (5) \(6.022 \times 10^{23}\) molecules (6) total mass of all atoms in 1 mole of a compound (7) smallest whole-number ratio of atoms present in a molecule (8) formula showing actual number of atoms present in a molecule (9) product formed when any carbon-containing compound is burned in \(\mathrm{O}_{2}\) (10) have the same empirical formulas, but different molecular formulas Column 2 (a) \(6.022 \times 10^{23}\) (b) atomic mass (c) mass of 1000 hydrogen atoms (d) benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) and acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\) (e) carbon dioxide (f) empirical formula (g) \(1.66 \times 10^{-24} \mathrm{~g}\) (h) molecular formula (i) molar mass (j) 1 mole
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