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University of Maine

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Answer

a. 4.82x10^3g, b. 556g, c. 0.56 moles, 3.4x10^22formula units, 6.8x10^22 lithium ions, 3.4x10^22sulfate ions, 3.4x10^22 sulfur atoms, 1.4x10^23 oxygen atoms

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## Discussion

## Video Transcript

When you're converting between quantities of given chemical substances, it's useful to draw out a roadmap or a plan to know how to get from the given information to the required quantity. So if, for example, you're given a quantity in moles 8.42 moles of a compound called chromium three sulfate Decca hydrate, and you want to convert that to however many grams or mass. The way to do that is by using the Moeller Mass. Remember, the molar mass is found by adding up the atomic masses of each element from the periodic table to given overall mass the compound. So the first step is to figure out what the chemical formula is for the substance. So chromium three greens. It's an ion with the CR with the three plus Sophie is s 04 to minus the Poly Atomic. I am Decca Hydrate means that it will have attached to it. 10 water molecules. The Decca is for 10 and hydrate means water. So the formula for this neutral compound is C R. Two s o for three 0.10 h 20 And we can find the molar mass of that using the periodic table two times the atomic mass of chromium, which is 51 0.996 plus sulfur. And there are three suffers, so three times 32.6 plus the oxygen's, of which there are 12 because there are four in the sulfate and there are three sulphate ions, 12 times 15 0.999 Remember, these numbers are the atomic weight on the periodic table plus hydrogen, and there are 10 times two or 20 times 1.8 And finally we have an additional 10 oxygen's from the water molecule, or one mole of this compound has a total mass of 572 0.3 grams. So if I have eight point for two moles change two grams, I multiply by 572.3, so 8.42 times 572.3 is equal to 4.82 times 10 to the third grams. You may be given a quantity and molecules rather than grant your moles, so we have 1.83 times 10 to the 24 molecules of die chlorine helped oxide, and we can change that two grams, but in order to do that, we have to change it to moles first. The relationship between molecules and mold is using off a god rose. Number one mole of any substance has 6.2 times 10 to the 23rd molecules. Once you've gone from molecules to moles, we can then use our molar Mass to find Gramps. So it's similar to what we did before. But we have one more step. So we need to know the formula of die Chlorine helped oxide di means to So it's C L two helped me in seven. So it's 07 and one mole. We'll have a mass of two times the atomic weight of chlorine 35.45 plus seven times the atomic weight of oxygen for 15.999 So one mole of die chlorine help dockside has a mass of 182.89 grams per mole. So if I start out with 1.83 times 10 to the 24 molecules, every one mole has 6.2 times 10 to the 23rd molecules and everyone mall also has a mass of 182 0.89 grams, so I multiply 1.83 times 10 to the 24th times 1 82.89 Your divide by 6.2 times 10 to the 23rd for a math of 556 grams. Another example you might see asks it to break down even further. If you have a given mass of a quantity 6.2 grams of a compound called lithium Sophie, we can find out how many moles that IHS using Mueller Mass. And once we have that information, we can find out how Maney Formula Units Formula Unit is just like a molecule. It's the base unit, the smallest particle of an Ionic compound. And so, again, in every one mole, there are 6.2 times 10 to the 23rd formula units we use of a God rose number. So the first step is to figure out what lithium self it is with the, um I on his ally plus and self, it is s o for two minus. So the compound is ally to s 04 and it's Moeller math, using the periodic table and adding up to lithium. One sulfur and four oxygen's equals 109 29 4 grams. We're starting out with 6.2 grams. Uh, ally to eso, for I can change that two moles by dividing by the molar Mass. Because I'm want moles. It's a 6.2 divided by 100 and 9.94 gives this 0.56 moles to find that and formula units multiply by Abu God. Rose number or 3.4 times 10 the 22nd formula units using the idea of what the formula unit is made up, up. We can also describe how Maney ions are present. So if I have 3.4 times 10 to the 22nd formula units in every formula unit, there are two lithium ions, or there would be a total of 6.8 times 10 to the 22nd with them I on. We can also describe how maney sulfate ions there would be because in looking at the formula in every formula unit, there's one sulfite I on for every formula unit or 3.4 times 10 to the 22 sulfate ion because also described how Maney Adams are present in every form really unit by looking at the ratio in the formula. In every formula unit, there is one sulfur atom based on the formula. So if I have 3.4 times 10 to the 22 formula, units have 3.4 times 10 to the 22 sulphur atoms. If instead I ask about oxygen atoms. When we look at the Formula Unit, we see that there are four oxygen atoms in every formula unit. So if I have 3.4 times 10 to the 22nd formula units, I should have four times as many oxygen atoms or 1.4 times 10 to the 23rd.

## Recommended Questions

Calculate each of the following quantities:

(a) Mass (g) of 8.35 mol of copper(I) carbonate

(b) Mass (g) of $4.04 \times 10^{20}$ molecules of dinitrogen pentoxide

(c) Amount (mol) and number of formula units in 78.9 $\mathrm{g}$ of sodium perchlorate

(d) Number of sodium ions, perchlorate ions, chlorine atoms, and oxygen atoms in the mass of compound in part (c)

Calculate the mass in grams of each of the following samples.

a. $1.27 \times 10^{-3}$ mol of carbon dioxide

b. $4.12 \times 10^{3}$ mol of nitrogen trichloride

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d. 18.0 mol of water

e. 62.7 mol of copper (II) sulfate

Calculate the mass in grams of each of the following samples.

a. 3.09 moles of ammonium carbonate

b. $4.01 \times 10^{-6}$ moles of sodium hydrogen carbonate

c. 88.02 moles of carbon dioxide

d. 1.29 mmol of silver nitrate

e. 0.0024 mole of chromium(III) chloride

Calculate each of the following quantities:

(a) Mass (g) of $6.44 \times 10^{-2} \mathrm{mol}$ of $\mathrm{MnSO}_{4}$

(b) Amount (mol) of compound in 15.8 $\mathrm{kg}$ of $\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}$

(c) Number of $\mathrm{N}$ atoms in 92.6 $\mathrm{mg}$ of $\mathrm{NH}_{4} \mathrm{NO}_{2}$

Calculate the number of moles of the indicated substance in each of the following samples.

a. 49.2 mg of sulfur trioxide

b. $7.44 \times 10^{4} \mathrm{kg}$ of lead(IV) oxide

c. 59.1 $\mathrm{g}$ of chloroform, CHCl_

d. 3.27 $\mathrm{mg}$ of trichloroethane, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}$

e. 4.01 $\mathrm{g}$ of lithium hydroxide

Calculate the mass in grams of each of the following samples.

a. $2.6 \times 10^{-2}$ moles of copper(II) sulfate, CuSO $_{4}$

b. $3.05 \times 10^{3}$ moles of tetrafluoroethylene, $\mathrm{C}_{2} \mathrm{F}_{4}$

c. 7.83 $\mathrm{mmol}(1 \mathrm{mmol}=0.001 \mathrm{mol})$ of $1,4$ -pentadiene, $\mathrm{C}_{5} \mathrm{H}_{8}$

d. 6.30 moles of bismuth trichloride, $\mathrm{BiCl}_{3}$

e. 12.2 moles of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$

Calculate the molar mass of each of the following compounds:

(a) $\mathrm{Fe}_{2} \mathrm{O}_{3},$ iron (III) oxide

(b) $\mathrm{BCl}_{3},$ boron trichloride

(c) $\left.\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}, \text { ascorbic acid (vitamin } \mathrm{C}\right)$

Calculate each of the following quantities:

(a) Total number of ions in 38.1 $\mathrm{g}$ of $\mathrm{SrF}_{2}$

(b) Mass $(\mathrm{kg})$ of 3.58 $\mathrm{mol}$ of $\mathrm{CuCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}$

(c) Mass $(\mathrm{mg})$ of $2.88 \times 10^{22}$ formula units of $\mathrm{Bi}\left(\mathrm{NO}_{3}\right)_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

Calculate the molar mass for each of the following substances.

a. ferrous sulfate

b. mercuric iodide

c. stannic oxide

d. cobaltous chloride

e. cupric nitrate

Calculate the number of moles of the indicated substance present in each of the following samples.

a. 1.28 $\mathrm{g}$ of iron(II) sulfate

b. 5.14 $\mathrm{mg}$ of mercury(II) iodide

c. 9.21$\mu \mathrm{g}$ of tin(IV) oxide

d. 1.26 $\mathrm{lb}$ of cobalt(II) chloride

e. 4.25 $\mathrm{g}$ of copper(II) nitrate