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What is the mass of oxygen in 148 grams of calcium hydroxide \(\left(\mathrm{Ca}(\mathrm{OH})_{2}\right)\) ? (A) 24 grams (B) 32 grams (C) 48 grams (D) 64 grams

Short Answer

Expert verified
The mass of oxygen in 148 grams of calcium hydroxide is approximately 64 grams (option D).

Step by step solution

01

Calculate the Molar Mass of Calcium Hydroxide

First, calculate the molar mass of calcium hydroxide. The molecular formula for calcium hydroxide is \(Ca(OH)_{2}\). The molar mass is calculated as follows: \(M_{Ca(OH)_{2}} = M_{Ca} + 2(M_{O} + M_{H}) = 40.08 g/mol + 2(16.00 g/mol + 1.01 g/mol) = 40.08 g/mol + 2(17.01 g/mol)= 74.1 g/mol\)
02

Calculate the Molar Mass of Oxygen in Calcium Hydroxide

Now determine the molar mass of the oxygen component. Note that there are two oxygen atoms in each molecule of calcium hydroxide: \(M_{O in Ca(OH)_{2}} = 2(16.00 g/mol) = 32.00 g/mol\)
03

Determine the Mass Proportion of Oxygen

Mass proportion of oxygen in calcium hydroxide is determined by dividing the molar mass of the oxygen component by the molar mass of the calcium hydroxide: Mass Proportion of Oxygen = \(\frac{M_{O in Ca(OH)_{2}}}{M_{Ca(OH)_{2}}} = \frac{32.00 g/mol}{74.1 g/mol} = 0.432\)
04

Activity the Mass Proportion of Oxygen

Compute the mass of oxygen in 148 grams of calcium hydroxide by multiplying the mass given by the proportion of oxygen: Mass of Oxygen = Mass Proportion of Oxygen * Mass of Calcium Hydroxide \(= 0.432 * 148 g = 63.93 g\)

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Key Concepts

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

chemistry problem solving
Chemistry problem solving often involves understanding how to manipulate data and perform calculations based on chemical formulas and reactions.
In this case, to solve the problem, the first step is to find the molar mass of the compound in question—calcium hydroxide (\( \mathrm{Ca(OH)_2} \)).
Understanding how each element contributes to the molar mass helps us accurately determine the composition of a compound.
For calculative purposes, knowing the atomic masses of the constituent elements, which are calcium (\( \mathrm{Ca} \)), oxygen (\( \mathrm{O} \)), and hydrogen (\( \mathrm{H} \)), from the periodic table is essential.
We then sum these weighted contributions to find the molar mass of the compound.
  • Calcium: 40.08 g/mol
  • Oxygen: 16.00 g/mol
  • Hydrogen: 1.01 g/mol
By adding these values appropriately as per the formula \( \mathrm{Ca(OH)_2} \), which includes two molecules of \( \mathrm{(OH)} \) groups, we derive the molar mass: 74.1 g/mol.
Performing calculations like these helps build fundamental problem-solving skills in chemistry.
mass proportion
Understanding the mass proportion is crucial in chemistry, especially when analyzing compounds.
Mass proportion in a compound refers to the ratio of the mass of one element to the total molar mass of the compound.
It's calculated by dividing the molar mass of the specific component by the entire compound's molar mass.
For calcium hydroxide:
  • Molar mass of calcium hydroxide: 74.1 g/mol
  • Molar mass of two oxygen atoms: 32.00 g/mol
To find the mass proportion of oxygen in calcium hydroxide, we compute:\[ \text{Mass proportion of oxygen} = \frac{32.00 \text{ g/mol}}{74.1 \text{ g/mol}} \approx 0.432 \]This means that approximately 43.2% of the mass of calcium hydroxide comes from oxygen.
Such calculations are important when determining the composition of compounds and are a fundamental tool in analytical chemistry.
molecular formula
A molecular formula provides the exact number and type of atoms in a single molecule of a compound.
For calcium hydroxide, the molecular formula \( \mathrm{Ca(OH)_{2}} \) indicates one calcium atom, two oxygen atoms, and two hydrogen atoms.
The molecular formula is like a recipe that shows the ratio of elements within the compound.
This helps chemists determine how molecules are structured and how they might react chemically with other substances.
From the molecular formula, we can derive the molar mass calculation, essential for determining the percentage by mass of each element.
  • It reveals how many hydrogen atoms interact with the oxygen to form two hydroxide groups (OH).
  • Helps calculate the total number of oxygen atoms, contributing to the overall oxygen mass.
Understanding how to read and interpret molecular formulas is a vital skill, especially in AP Chemistry, for developing solutions to complex chemical problems.
AP Chemistry preparation
As you prepare for AP Chemistry, mastering concepts like molar mass calculation and mass proportion will be essential.
These topics form the backbone of chemical understanding, aiding in experiments and theoretical assessments.
Being proficient in interpreting molecular formulas enhances your ability to conduct accurate stoichiometric calculations. Here are key focus points:
  • Effective use of the periodic table to extract necessary atomic masses.
  • Skillful calculation of molar masses and mass proportions to analyze various compounds.
  • Accurate interpretation of molecular formulas to understand compound composition.
AP Chemistry often involves applying these simple mathematical skills to complex problems, such as finding impurities or determining reactant quantities in reactions.
Practice regularly will make these concepts second nature, building your confidence in tackling even the most challenging chemistry questions.

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Most popular questions from this chapter

Why does an ion of phosphorus, \(\mathrm{P}^{3-}\) , have a larger radius than a neutral atom of phosphorus? (A) There is a greater Coulombic attraction between the nucleus and the electrons in \(\mathrm{P}^{3}\) . (B) The core electrons in \(\mathrm{P}^{3-}\) exert a weaker shielding force than those of a neutral atom. (C) The nuclear charge is weaker in \(\mathrm{P}^{3-}\) than it is in P. (D) The electrons in \(\mathrm{P}^{3-}\) have a greater Coulombic repulsion than those in the neutral atom.

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A 22.0 gram sample of an unknown gas occupies 11.2 liters at standard temperature and pressure. Which of the following could be the identity of the gas? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{SO}_{3}\) (C) \(\mathrm{O}_{2}\) (D) He

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

A mixture of helium and neon gases has a total pressure of 1.2 atm. If the mixture contains twice as many moles of helium as neon, what is the partial pressure due to neon? (A) 0.2 atm (B) 0.3 atm (C) 0.4 atm (D) 0.8 atm

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