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Chapter 3: Problem 17

Calculate each of the following quantities: (a) Total number of ions in \(38.1 \mathrm{~g}\) of \(\mathrm{SrF}_{2}\) (b) Mass (kg) of \(3.58 \mathrm{~mol}\) of \(\mathrm{CuCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) Mass (mg) of \(2.88 \times 10^{22}\) formula units of \(\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
(a) 5.47 \times 10^{23} \text{ions}(b) 0.612 \text{kg}(c) 23200 \text{mg}

Step by step solution

01

Calculate moles of \(\text{SrF}_2\) in 38.1 g

First, find the molar mass of \(\text{SrF}_2\). \(\text{Molar mass of SrF}_2 = \text{molar mass of Sr} + 2 \times \text{molar mass of F} = 87.62 \mathrm{g/mol} + 2 \times 19.00 \mathrm{g/mol} = 125.62 \mathrm{g/mol}\)Then, calculate the moles of \(\text{SrF}_2\) using the formula: \(\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{38.1 \mathrm{g}}{125.62 \mathrm{g/mol}} = 0.303 \text{mol}\)
02

Calculate total ions in \(\text{SrF}_2\)

Each formula unit of \(\text{SrF}_2\) dissociates into 3 ions (1 Sr²⁺ ion and 2 F⁻ ions). Thus, the total number of ions can be calculated as: \(\text{Total ions} = 0.303 \text{mol} \times 3 \times 6.022 \times 10^{23} \text{ions/mol} = 5.47 \times 10^{23} \text{ions}\)
03

Find molar mass of \(\text{CuCl}_2 \cdot 2 \text{H}_2\text{O}\)

Calculate the molar mass of \(\text{CuCl}_2 \cdot 2 \text{H}_2\text{O}\).\(\text{Molar mass of Cu} = 63.55 \mathrm{g/mol}\), \(\text{Cl}_2 = 2 \times 35.45 \mathrm{g/mol}\), \(\text{H}_2\text{O} = 2 \times 18.02 \mathrm{g/mol}\).\(\text{molar mass of CuCl}_2 \cdot 2 \text{H}_2\text{O} = 63.55 + (2 \times 35.45) + (2 \times 18.02) = 170.99 \mathrm{g/mol}\)
04

Calculate mass of \(\text{CuCl}_2 \cdot 2 \text{H}_2\text{O}\)

Using the moles given, calculate the mass: \(\text{Mass} = \text{moles} \times \text{molar mass} = 3.58 \text{mol} \times 170.99 \mathrm{g/mol} = 612 \mathrm{g} = 0.612 \mathrm{kg}\)
05

Calculate moles of \(\text{Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O}\)

First, calculate the moles of \(\text{Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O}\): \(\text{Moles} = \frac{2.88 \times 10^{22} \text{formula units}}{6.022 \times 10^{23} \text{formula units/mol}} = 0.0478 \text{mol}\)
06

Find molar mass of \(\text{Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O}\)

Calculate the molar mass:\(\text{Molar mass of Bi} = 208.98 \mathrm{g/mol}\), \(\text{NO}_3 = 3 \times (14.01 + 3 \times 16.00) = 3 \times 62.01 = 186.03 \mathrm{g/mol}\), \(\text{H}_2\text{O} = 5 \times 18.02 = 90.1 \mathrm{g/mol}\).\(\text{Molar mass of Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O} = 208.98 + 186.03 + 90.1 = 485.11 \mathrm{g/mol}\)
07

Calculate mass of \(\text{Bi(NO}_3)_3 \cdot 5 \text{H}_2\text{O}\)

Using the moles calculated, find the mass: \(\text{Mass} = \text{moles} \times \text{molar mass} = 0.0478 \text{mol} \times 485.11 \mathrm{g/mol} = 23.2 \mathrm{g} = 23200 \mathrm{mg} \)

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is critical in chemistry. The molar mass is the mass of one mole of a substance, often in grams per mole (g/mol). This involves summing the atomic masses of all atoms in a molecule. Consider \(\text{SrF}_2\). First, find the atomic masses: Strontium (Sr) is 87.62 g/mol, and Fluorine (F) is 19.00 g/mol. Because there are two fluorine atoms, double their mass and add it to the strontium's mass: 87.62 + 2 * 19.00 = 125.62 g/mol. Repeat this process for any compound to find its molar mass.
Moles to Mass Conversion
Once you have the molar mass, converting moles to mass is straightforward. Use the formula: \(\text{Mass} = \text{moles} \times \text{molar mass}\). For example, let's find the mass of \(\text{CuCl}_2 \cdot 2 \text{H}_2\text{O}\) if you have 3.58 mol. First, calculate its molar mass: Copper (Cu) is 63.55 g/mol, Chlorine (Cl) is 35.45 g/mol, and Water (H_2O) is 18.02 g/mol. So, \(\text{63.55} + 2 \times \text{35.45} + 2 \times \text{18.02} = 170.99 \text{g/mol}\). Multiply this by the moles: 3.58 mol \times 170.99 \text{g/mol} = 612 g or 0.612 kg. Remember to always use the same units throughout your calculations.
Dissociation of Ionic Compounds
Ionic compounds, like \(\text{SrF}_2\), dissociate in water into their respective ions. Each formula unit of \(\text{SrF}_2\) dissociates into one \(\text{Sr}^{2+}\) ion and two \(\text{F}^{-}\) ions. This total break down results in three ions from one formula unit. To find the number of ions, use the number of moles and Avogadro's number. For example, 0.303 mol of \(\text{SrF}_2\) gives: \(\text{0.303} \times 3 \times 6.022 \times 10^{23} = 5.47 \times 10^{23}\) ions. Note the use of Avogadro's number (6.022 \times 10^{23}) here to convert from moles to individual ions.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry, roughly \(\text{6.022} \times 10^{23}\). It represents the number of units (atoms, molecules, ions, etc.) in one mole of any substance, a link between macroscopic and microscopic worlds. For instance, if you have \(\text{2.88} \times 10^{22}\) formula units of \(\text{Bi(NO_3)_3 \cdot 5 \text{H}_2O}\), you can find the number of moles by dividing by Avogadro's number: \(\frac{\text{2.88} \times 10^{22}}{6.022 \times 10^{23}} = 0.0478 \text{moles}\). This simple division brings the complex world of molecules into an accessible quantity that can be easily worked with and measured.

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Most popular questions from this chapter

Fluorine is so reactive that it forms compounds with several of the noble gases. (a) When \(0.327 \mathrm{~g}\) of platinum is heated in fluorine, \(0.519 \mathrm{~g}\) of a dark red, volatile solid forms. What is its empirical formula? (b) When \(0.265 \mathrm{~g}\) of this red solid reacts with excess xenon gas, \(0.378 \mathrm{~g}\) of an orange-yellow solid forms. What is the empirical formula of this compound, the first to contain a noble gas? (c) Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra- and hexafluorides, \(1.85 \times 10^{-4} \mathrm{~mol}\) of xenon reacted with \(5.00 \times 10^{-4} \mathrm{~mol}\) of fluorine, and \(9.00 \times 10^{-6} \mathrm{~mol}\) of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide: $$ 4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g) $$ When 1.82 mol of \(\mathrm{HCl}\) reacts with excess \(\mathrm{MnO}_{2}\), how many (a) moles of \(\mathrm{Cl}_{2}\) and (b) grams of \(\mathrm{Cl}_{2}\) form?

Calculate the molar mass of each of the following: (a) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}\) (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{BrF}_{3}\)

Find the empirical formula of each of the following compounds: (a) \(0.039 \mathrm{~mol}\) of iron atoms combined with \(0.052 \mathrm{~mol}\) of oxygen atoms; (b) \(0.903 \mathrm{~g}\) of phosphorus combined with \(6.99 \mathrm{~g}\) of bromine; (c) a hydrocarbon with 79.9 mass \% carbon

Oxygen is required for the metabolic combustion of foods. Calculate the number of atoms in \(38.0 \mathrm{~g}\) of oxygen gas, the amount absorbed from the lungs in about 15 min when a person is at rest.
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