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Problem 120

Elements $X(\text {green})$ and $\mathrm{Y}$ (pur…

01:20
University of Maine
Problem 119

Calculate each of the following quantities:
(a) Amount (mol) of 0.588 g of ammonium bromide
(b) Number of potassium ions in 88.5 $\mathrm{g}$ of potassium nitrate
(c) Mass (g) of 5.85 mol of glycerol $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)$
(d) Volume (L) of 2.85 $\mathrm{mol}$ of chloroform $\left(\mathrm{CHCl}_{3} ; d=1.48 \mathrm{g} / \mathrm{mL}\right)$
(e) Number of sodium ions in 2.11 mol of sodium caa
(f) Number of atoms in 25.0$\mu \mathrm{g}$ of cadmium
(g) Number of atoms in 0.0015 mol of fluorine gas

Answer

a) $\mathrm{n}\left(\mathrm{NH}_{4} \mathrm{Br}\right)=0,006 \mathrm{mol}$
b) $\mathrm{N}\left(\mathrm{KNO}_{3}\right)=\mathrm{n}^{*} \mathrm{N}_{A}=5,27^{*} 10^{23}$ molecules
c) $\mathrm{m}=538,73 \mathrm{g}$
d) $\mathrm{V}=229,89 \mathrm{mL}$
e) $\mathrm{N}(\mathrm{Na})=2,54^{*} 10^{24}$ ions
f) $\mathrm{N}(\mathrm{Cd})=1,34^{*} 10^{17}$ atoms
g) $\mathrm{N}(\mathrm{F})=\mathrm{n}(\mathrm{F})^{*} \mathrm{N}_{A}=1,81^{*} 10^{21}$ atoms


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Video Transcript

being able to convert toe in given chemical quantities involves using several different understandings, including of a God rose number, the molar mass and finally an understanding of the chemical formula. So here are some examples of conversions you should be able to dio. If we have 0.5 88 grams of a substance called ammonium bromide, we can convert that two moles of the same substance. The first step is to know the formula of ammonium bro mind. This is an ionic compound made up of the NH four cat I in with the plus one charge and the BR minus, and I am so the formula is en age for BR. If we have grams and we want to go to moles, we're going to use the molar mass. So to find that we use our periodic table and we add up the molar masses of each of the individual elements. So nitrogen has a molar mass of 14.7 plus four times the molar mass of hydrogen, plus the molar mass of grooming or one more equals 97 0.943 grams so we can use our given quantity and divide by the molar Mass to get the number of moles. Another example looks at having a certain compound 88.5 grams of a compound called potassium nitrate, and we want to know how many ions are in this quantity. Potassium nitrate is an ionic compound made up of two ions K plus in the N 03 minus, so 1 to 1 ratios. Which formula is Kono three. But it's made up of two ions for every formula unit. So in this case will have our mass in grams That will need to change two moles that will then need to change to formula units and then finally, to the number of ions of potassium. So we start with our given Mass, which we divide by the molar Mass, which I found just like I did in the previous example. By adding up the molar mass of K and plus three Oxygen's and then to change to formula units. I use of a God rose number, and finally, I know that there's one potassium ion for every formula unit, so I multiply 6.2 times, 10 to the 23rd times, 88.5 and then divide by 101.102 or they're 5.27 times 10 to the 23rd. Potassium ions. If you're given moles of a substance, you confined grams. And again, I'm going to use the molar Mass. So to find the Moeller math, I'm going to take three times. The molar mass of carbon was eight times the molar mass of hydrogen, plus three times the molar mass of oxygen or one mole of this compound has a molar mass of 92 0.94 grams. So if I have 5.85 moles to change it to grams, I simply multiply by the molar mass or 500 and 39 grams. Sometimes you're given a quantity and ask of a liquid and asked to find its volume. So if I have 2.85 Knowles of a substance ch C. L three and I want to find its volume and leaders, I also need to know the density. So I'll go through starting off with my moles, changing it to mass by using the molar mass and then changing it to volume using the density grams per milliliter. We'll use the molar Mass and then the density. So I find the molar mass justifies. I've done in the other examples by adding up the individual molar masses and I multiply by the molar mass. See that this quantity is equivalent to 340 grams. I also know that density is mass divided by volume and I can solve For Mal you. Volume is 230 milliliters, which in leaders divide by 1000 his 0.230 leaders. Another example Looking at ionic compounds is if I have 2.11 moles of a compound called sodium carbonate, then I want to know how many sodium ions. So I need to know the formula. Sodium carbonate. It's made up of the N A plus ion and the Poly Atomic Ion CEO three to minus because there's a plus one and a two minus. The formula is an A to C 03 That means there are two sodium ions for every formula unit, so I have moles. Then I'll change to formula units using over God rose number and then the number of ions from the formula. So 2.11 Moles times 6.2 times, 10 to the 23rd for every mole times to sodium ion for every formula unit or 2.54 times 10 to the 24. If I am given a quantity oven element, I confined. How many atoms there are 25 0.0 micrograms of cadmium is how many Adams. First, we have to change two grams. So remember that a microgram is 10 to the minus six of a gram, so this mass is equivalent to 2.50 times 10 to the minus five grams. Once they have mass, Ingram's I can convert it to moles using Moeller math. And then I confined Adams by using our God rose number. I look up on the periodic table, the molar mass of cadmium. Since it's an element, it's just the number on the periodic table 112 0.41 grams. So dividing my initial math by the molar Mass and then I multiplied by Aava God Rose number and find that this is equivalent to 1.34 times 10 to the 17. Adam in the final example gives me a quantity of flooring gaffe and asks how Maney Adams. So I have mold and Adam, so it's fairly straightforward. The only trick is remembering that Florian gases f two, not just f. So there will be two atoms of Florence and every molecule of Florida. So I start off with my given quantity and moles changed to molecules using over God rose number. And then I know that there are two Adams for every one molecule for 1.8 times 10 to the 22nd Adams.

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