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Ch.2 - Atoms, Molecules & Ions
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.2 - Atoms, Molecules & IonsProblem 81
Chapter 2, Problem 81

In borane, one part hydrogen combines with 3.6 parts boron by mass. A compound containing only hydrogen and boron contains 6.0 g of hydrogen and 43.2 g of boron. Could this compound be borane? If it is not borane, show that the law of multiple proportions is followed for borane and this other substance.

Verified step by step guidance
1
Calculate the mass ratio of boron to hydrogen in the given compound by dividing the mass of boron by the mass of hydrogen: \( \text{Mass ratio} = \frac{43.2 \text{ g B}}{6.0 \text{ g H}} \).
Compare the calculated mass ratio from the compound to the mass ratio given for borane (3.6 parts boron to 1 part hydrogen).
If the mass ratio from the compound matches the mass ratio for borane, then the compound could be borane. If not, proceed to the next step.
To show the law of multiple proportions, calculate the ratio of the two mass ratios (the one from the compound and the one for borane).
If the ratio of the two mass ratios is a simple whole number, then the law of multiple proportions is followed, indicating that the compound is a different boron-hydrogen compound.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Law of Definite Proportions

The Law of Definite Proportions states that a chemical compound always contains its component elements in fixed ratio by mass, regardless of the sample size or source. In the case of borane, the mass ratio of hydrogen to boron is consistent, which is essential for identifying whether a given sample can be classified as borane.

Law of Multiple Proportions

The Law of Multiple Proportions asserts that when two elements form more than one compound, the ratios of the masses of one element that combine with a fixed mass of the other element can be expressed as small whole numbers. This principle can be applied to compare borane with another hydrogen-boron compound to demonstrate that they adhere to this law.
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Molar Mass and Stoichiometry

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. Understanding stoichiometry allows us to calculate the mass ratios of elements in compounds. By determining the molar masses of hydrogen and boron, we can analyze the mass ratios in the given compound and compare them to those in borane to assess whether they are consistent.
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Related Practice
Textbook Question
The atomic weight of carbon (12.011) is approximately 12 times that of hydrogen (1.008).(a) Show how you can use this knowledge to calculate pos-sible formulas for benzene, ethane, and ethylene (Prob-lem 2.82).
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In methane, one part hydrogen combines with three parts carbon by mass. If a sample of a compound containing only carbon and hydrogen contains 32.0 g of carbon and 8.0 g of hydrogen, could the sample be methane? If the sample is not methane, show that the law of multiple proportions is fol-lowed for methane and this other substance.
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Benzene, ethane, and ethylene are just three of a large num-ber of hydrocarbons—compounds that contain only carbon and hydrogen. Show how the following data are consistent with the law of multiple proportions.
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A sample of mercury with a mass of 114.0 g was combined with 12.8 g of oxygen gas, and the resulting reaction gave 123.1 g of mercury(II) oxide. How much oxygen was left over after the reaction was complete?
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A sample of CaCO3 was heated, causing it to form CaO and CO2 gas. Solid CaO remained behind, while the CO2 escaped to the atmosphere. If the CaCO3 weighed 612 g and the CaO weighed 343 g, how many grams of CO2 were formed in the reaction?
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