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Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples. a. 2.17 moles of germanium, Ge b. \(4.24 \mathrm{mmol}\) of lead(II) chloride ( \(1 \mathrm{mmol}=1 / 1000 \mathrm{~mol}\) ) c. 0.0971 mole of ammonia, \(\mathrm{NH}_{3}\) d. \(4.26 \times 10^{3}\) moles of hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\) e. 1.71 moles of iodine monochloride, \(1 \mathrm{C}=\)

Short Answer

Expert verified
a. The mass of 2.17 moles of germanium (Ge) is 157.50 g. b. The mass of 4.24 mmol of lead(II) chloride is 1.18 g. c. The mass of 0.0971 mole of ammonia, NH3, is 1.65 g. d. The mass of 4.26 × 10^3 moles of hexane, C6H14, is 367,000 g. e. The mass of 1.71 moles of iodine monochloride, ICl, is 277.62 g.

Step by step solution

01

Find the molar mass of germanium (Ge)

The molar mass of germanium (Ge) is 72.63 g/mol.
02

Calculate the mass in grams

Now, multiply the moles of Ge (2.17 mol) by its molar mass (72.63 g/mol): Mass = 2.17 mol × 72.63 g/mol = 157.50 g b. 4.24 mmol of lead(II) chloride (1 mmol = 1/1000 mol)
03

Find the molar mass of lead(II) chloride

The molar mass of lead(II) chloride is PbCl2 = (207.2+ 2×35.45) g/mol = 278.10 g/mol.
04

Convert millimoles to moles and calculate the mass in grams

Convert 4.24 mmol to moles: 4.24 mmol × (1 mol / 1000 mmol) = 0.00424 mol Now, multiply the moles of PbCl2 (0.00424 mol) by its molar mass (278.10 g/mol): Mass = 0.00424 mol × 278.10 g/mol = 1.18 g c. 0.0971 mole of ammonia, NH3
05

Find the molar mass of ammonia (NH3)

The molar mass of ammonia (NH3) is (14.01 + 3×1.008) g/mol = 17.034 g/mol.
06

Calculate the mass in grams

Multiply the moles of NH3 (0.0971 mol) by its molar mass (17.034 g/mol): Mass = 0.0971 mol × 17.034 g/mol = 1.65 g d. 4.26 × 10^3 moles of hexane, C6H14
07

Find the molar mass of hexane (C6H14)

The molar mass of hexane (C6H14) is (6×12.01 + 14×1.008) g/mol = 86.178 g/mol.
08

Calculate the mass in grams

Multiply the moles of C6H14 (4.26 × 10^3 mol) by its molar mass (86.178 g/mol): Mass = 4.26 × 10^3 mol × 86.178 g/mol = 367000 g e. 1.71 moles of iodine monochloride, ICl
09

Find the molar mass of iodine monochloride (ICl)

The molar mass of iodine monochloride (ICl) is (126.9 + 35.45) g/mol = 162.35 g/mol.
10

Calculate the mass in grams

Multiply the moles of ICl (1.71 mol) by its molar mass (162.35 g/mol): Mass = 1.71 mol × 162.35 g/mol = 277.62 g

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Key Concepts

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

Average Atomic Mass
Understanding the concept of average atomic mass is essential when dealing with various elements and compounds. The average atomic mass of an element is the weighted average of the masses of all its isotopes. Isotopes are different forms of the same element, each having a different number of neutrons and, therefore, a different mass.

The average atomic mass takes into account the abundance of each isotope in nature. You can find the average atomic mass on the periodic table for each element. It’s usually expressed in atomic mass units (amu). For example, the average atomic mass of germanium (\( \text{Ge} \)) is listed as approximately 72.63 amu. This value is calculated based on the natural abundance and mass of germanium isotopes.

This concept helps us determine the molar mass of an element, which is essential in converting between moles and grams. By knowing the average atomic mass, you can calculate how much one mole of an element weighs in grams. For instance, 1 mole of germanium weighs 72.63 grams.
Moles to Grams Conversion
The conversion from moles to grams is a fundamental chemical calculation, necessary for understanding the mass of a sample. When you're given moles and need to find grams, you will use the molar mass. The molar mass is the mass of 1 mole of a substance, measured in grams per mole (g/mol) and is equivalent to the element’s average atomic or molecular mass.

To convert moles to grams, you multiply the number of moles by the molar mass:

  • First, determine the molar mass of the substance. Use the periodic table or provided data for this, such as 86.178 g/mol for hexane (\( \text{C}_6\text{H}_{14} \)).
  • Next, multiply the given number of moles by the molar mass to find the mass in grams. For example, if dealing with hexane, for 4.26 × 10^3 moles, the mass is calculated as:
\[\text{Mass} = 4.26 \times 10^3 \text{ moles} \times 86.178 \text{ g/mol} = 367,000 \text{ g}\]

This conversion is crucial when working in laboratories or solving chemical equations, as it helps in quantifying the actual substance you have based on its mole count.
Molecular Compounds
Molecular compounds are formed when two or more elements share electrons, creating covalent bonds. They typically consist of nonmetals and are characterized by distinct properties. Understanding molecular compounds is critical when calculating their molar masses and, thus, their use in chemical equations and reactions.

Each molecular compound has a specific chemical formula representing the elements present and their atomic ratio. For instance, ammonia (\( \text{NH}_3 \)) consists of one nitrogen (N) atom and three hydrogen (H) atoms. The molar mass of molecular compounds is calculated by adding up the average atomic masses of all the atoms in the molecule.

To find the molar mass of ammonia:
  • Determine the atomic mass of nitrogen: 14.01 amu, and hydrogen: 1.008 amu.
  • Multiply the atomic mass of each element by its number of atoms in the molecule. Add these values together: \(14.01 + 3 \times 1.008 = 17.034 \text{ g/mol}\).
This information is key for scientists to understand chemical reactions, as it allows them to correctly balance equations and calculate reactant and product quantities in molecular compounds.

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Most popular questions from this chapter

Consider the following unbalanced chemical equation. $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(67.4 \mathrm{~g}\) of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

Your text talks about several sorts of "yield" when experiments are performed in the laboratory. Students often confuse these terms. Define, compare, and contrast what are meant by theoretical yield, actual yield, and percent yield.

Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst. $$ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s) $$ What is the theoretical mass of xenon tetrafluoride that should form when \(130 .\) g of xenon is reacted with 100.8 of \(\mathrm{F}_{2} ?\) What is the percent yield if only \(145 \mathrm{~g}\) of \(\mathrm{XeF}_{4}\) is actually isolated?

For each of the following unbalanced equations, calculate how many moles of the second reactant would be required to react completely with 0.413 moles of the first reactant. a. \(\operatorname{Co}(s)+\mathbf{F}_{2}(g) \rightarrow \operatorname{CoF}_{3}(s)\) b. \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{H}_{2}(g)\) d. \(\mathrm{Cu}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(s)\)

For cach of the following reactions, give the balanced cquation for the reaction and state the meaning of the equation in terms of the numbers of individual molecules and in terms of moles of molecules. a. \(\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(g)\) b. \(\mathrm{XeF}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Xe}(g)+\mathrm{HF}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NO}_{2}(g)\) d. \(\mathrm{NaHSO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\)

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