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Chapter 10: Problem 17

Determine the number of moles in each of the following. a. 25.5 g Ag b. 300.0 g S

Short Answer

Expert verified
For the given exercise: a. There are \(0.236 \, mol\) of Ag in 25.5 g of Ag b. There are \(9.35 \, mol\) of S in 300.0 g of S

Step by step solution

01

Find the molar mass of Ag (Silver)

To find the molar mass of silver (Ag), we can look at the periodic table. The molar mass of Ag is approximately 107.87 g/mol.
02

Calculate the number of moles of Ag

Now we can use the formula to find the number of moles of silver in 25.5 g of Ag. Number of moles = mass of substance (g) / molar mass of the substance (g/mol) Number of moles of Ag = \( \frac{25.5 \, g}{107.87 \, g/mol} = 0.236 \, mol \) There are 0.236 moles of silver (Ag) in 25.5 g of Ag.
03

Find the molar mass of S (Sulfur)

To find the molar mass of sulfur (S), we can look at the periodic table. The molar mass of S is approximately 32.07 g/mol.
04

Calculate the number of moles of S

Now we can use the formula to find the number of moles of sulfur in 300.0 g of S. Number of moles = mass of substance (g) / molar mass of the substance (g/mol) Number of moles of S = \( \frac{300.0 \, g}{32.07 \, g/mol} = 9.35 \, mol \) There are 9.35 moles of sulfur (S) in 300.0 g of S. So, for the given exercise: a. There are 0.236 moles in 25.5 g of Ag b. There are 9.35 moles in 300.0 g of S

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Key Concepts

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

Molar Mass
Understanding molar mass is essential when performing mole calculations. The molar mass is the weight of one mole of an element or compound, typically expressed in grams per mole (\( \text{g/mol} \)). To find the molar mass, you can use the periodic table, which provides the atomic mass of each element. The atomic mass, usually listed below the element symbol, directly represents the molar mass of that particular element.

For example:
  • Silver (Ag) has a molar mass of approximately 107.87 g/mol.
  • Sulfur (S) has a molar mass of approximately 32.07 g/mol.
This property allows you to convert between grams and moles, providing a bridge for converting mass to moles, aiding in the calculation of amounts in chemical reactions.
Stoichiometry
Stoichiometry is the study of the quantitative relationships in chemical reactions. It allows chemists to predict how much of each substance is needed or produced in a reaction. Mole calculations are central to stoichiometry, as they help determine the proportions between reactants and products.

To utilize stoichiometry effectively:
  • Kep in mind the balanced chemical equation, which shows the relation between moles of reactants and products.
  • Use mole ratios derived from the balanced equation to convert from moles of one substance to moles of another.
The mole concept ensures that elements combine in certain proportions and provides insight into the amounts needed for reactions, assisting in predicting reaction yields and optimizing processes.
Chemical Formulas
Chemical formulas represent the elements in a compound and the number of each element's atoms. They are fundamental to understanding the composition of molecules and compounds. A simple chemical formula gives you a direct way to calculate the molar mass of compounds.

Let's consider a chemical formula like \( \text{H}_2 \text{O} \) (water), which tells us:
  • 2 atoms of hydrogen (H) - each with an atomic mass of approximately 1.01 g/mol.
  • 1 atom of oxygen (O) - with an atomic mass of approximately 16.00 g/mol.
To find the molar mass of water, you would add up these values: \( 2 \times 1.01 + 16.00 = 18.02 \ ext{g/mol}\).

Understanding chemical formulas allows us to analyze the amount of different elements in a compound and serves as a tool for calculations in chemical reactions, elucidating both stoichiometry and molar mass.

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Most popular questions from this chapter

How many molecules are contained in each compound? a. 1.35 mol of carbon disulfide (CS_{2} ) b. 0.254 mol of diarsenic trioxide \(\left(\mathrm{As}_{2} \mathrm{O}_{3}\right)\) c. 1.25 mol of water d. 150.0 mol of \(\mathrm{HCl}\)

Chromium(III) nitrate forms a hydrate that is 40.50% water by mass. What is its chemical formula?

Determine the mass in grams of each element. a. \(1.33 \times 10^{22} \mathrm{mol}\) of \(\mathrm{Sb}\) b. \(4.75 \times 10^{14} \mathrm{mol}\) of \(\mathrm{Pt}\) c. \(1.22 \times 10^{23} \mathrm{mol}\) of Ag d. \(9.85 \times 10^{24} \mathrm{mol}\) of Cr

Which has more atoms, 10.0 g of C or 10.0 g of Ca? How many atoms does each have?

Express each answer with the correct number of significant figures. (Chapter 2) a. \(18.23-456.7\) b. \(4.233 \div 0.0131\) c. \((82.44 \times 4.92)+0.125\)
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