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Problem 19

Calculate each of the following quantities: (a) …

12:47
University of Maine
Problem 18

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)

Answer

a) $\mathrm{m}\left(\mathrm{Cu}_{2} \mathrm{CO}_{3}\right)=1562,37 \mathrm{g}$
b) $\mathrm{m}=0,0724 \mathrm{g}$
c) $\mathrm{n}=0,644 \mathrm{mol}$ $\mathrm{N}=3,88^{*} 10^{23}$
d) $\mathrm{N}\left(\mathrm{Na}^{+}\right)=\mathrm{N}\left(\mathrm{NaClO}_{4}\right)=3,88^{*} 10^{23}$ ions
$\mathrm{N}\left(\mathrm{ClO}_{4}-\right)=\mathrm{N}\left(\mathrm{NaClO}_{4}\right)=3,88^{*} 10^{23} \mathrm{ions}$
$\mathrm{N}(\mathrm{Cl})=\mathrm{N}\left(\mathrm{NaClO}_{4}\right)=3,88^{*} 10^{23}$ atoms
$\mathrm{N}(\mathrm{O})=4^{*} \mathrm{N}\left(\mathrm{NaClO}_{4}\right)=4^{*} 3,88^{*} 10^{23}=1,55^{*} 10^{24}$ atoms


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Video Transcript

when we're converting between given quantities of measurement describing amounts of substances it's helpful to draw out a roadmap were planned to know which conversion to use when. So if, for example, we're starting off with a given quantity of Yuri's in this case 8.35 moules of copper, one carbonate. Okay, and we want to know the mass equivalent or how many grams to go between moles and mass. The relationship that we use is the molar mass, and we find the molar mass by using the formula of the compound and the periodic table. So to figure out what copper carbonate is copper is See you in the room. Number one indicates it's got a charge plus one. The carbonate ion is CEO three tu minus. So to make it a neutral ionic compound, the formula is see you too, CEO three. And then using our periodic table, we find the molar. Mass. One mole is equal to the mass of copper from the periodic table times two so two times 63 0.5 46 plus the massive carbon 12.11 plus three times the mass of oxygen. So these numbers here have all come from the periodic table is the atomic mass of each element. And so when we multiply and add them together, one mole of copper Kirk Copper carbonate has a molar mass of 187 0.10 grams. So using this relationship, the molar mass. If we have 8.35 moles, we can change it to grams by multiplying by the molar mass. So we multiply 8.35 times 187.10 or a molar mass of a total mass of 1.56 times 10 to the third grams. So 8.35 moles is equal to 1.56 times 10 to the third cramps. Another example. We might be given molecules instead of molds when we have molecules. The relationship that we used to change two moles is Allah God rose number because that tells us that in every mole there are 6.2 times 10 to the 23rd molecules. And then if we want to go from molecules two grams after we changed two moles, we again use our molar mass. So in this case, We're talking about a compound called Die Nitrogen hence oxide. So to find the molar mass of die nitrogen pent oxide, we find the formula. We know that it has nitrogen and the dye means there are two. It has oxygen and the pent means five. So I have n 205 So we confined our Moeller Mass just like we did above by multiplying two times the molar mass of nitrogen were 14 point 007 plus five times the molar mass of oxygen 15.999 We're one more is equivalent to 108 0.1 grams. This is a two step problem where we first start off with our molecules. We changed two moles using of a god rose number one more is equal to 6.2 times 10 to the 23rd molecules. And then we changed two grams using our Mueller Mass. We're one more has a mass of 108.1 grams, so multiplying 4.4 times 10 to the 20 times 108.1 women dividing by 6.2 times 10 to the 23rd gives us the answer of 0.0 7 to 5 grams. You may be given a mass to start with and then find quantities and moles, formula units and then even smaller descriptions based on what the chemical formula is. So if we have 78.9 grams of a compound called sodium per chlorate, we can find how many moles this is equivalent to and again grams to moles. We use the molar mass. We confined on a smaller level. How Maney formula units this is equivalent to, and a formula unit is just the smallest unit of an Ionic compound. So again, we use Allah God Rose number. So before we start converting, we need to figure out what sodium per chlorate is. Sodium is an eye on an A plus. The per chlorate ion is C L 04 with an overall minus one charge. So the formula is N a. C l 04 and the molar mass is equal to the mass of sodium plus the massive chlorine plus four oxygen's or 122.44 grams so I can convert from my given grams, two moles because I want it in moles I divide by the Moeller math. So 78.9 divided by 100 and 22.44 in this equals 0.644 moles find the equivalent and formula units. I take my number of moles and I multiply by Abu God Rose number So this equals 3.88 times 10 to the 23rd formula units Ah, an a c l o For we can use that information to break it down even further If I want to know how maney sodium ions there are for every formula unit of an a c l 04 there is one sodium ion. So if I have 3.88 times 10 to the 23rd formula unit, I will also have the same amount of sodium ion. Similarly, for every formula unit, I have one per chlorate ion. So I should also have 3.88 times 10 to the 23rd quarry. I am for every same quantity of formula units. If I want to know how many chlorine atoms, if I look at the formula here in this formula for every one of these, there's one chlorine Adam. So if I have 3.88 times 10 to the 23rd formula units, I should have the same amount of chlorine atoms or 3.88 times 10 to the 23rd Chlorine atoms. If, however, I ask how maney oxygen atoms are, you can see that in one formula unit there are four oxygen atoms. So that means that there will be four times as many oxygen atoms in the given quantity. Okay, who are 1.55 times 10 to the 24 atoms of oxygen.

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