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Problem 10 Hard Difficulty

Find the molar mass.
a. $\mathrm{CsCl}$
b. $\mathrm{KClO}_{3}$
c. $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$
d. $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$
e. $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}$

Answer

a) 1(132.9)+1(35.45)=168.35 g/mol
b) 1(39.1)+1(35.45)+3(16)=122.55 g/mol
c) 6(12.01)+12(1.01)+6(16)=180.18 g/mol
d) 2(14.01)+8(1.01)+1(1.01)+1(30.97)+4(16)= 132.08 g/mol
e) 2(12.01)+5(1.01)+1(14.01)+2(16)=75.08 g/mol

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Video Transcript

to calculate the molar mass of these compounds. We simply need to recognize how many moles of each element that we have in a mole of the compound and then use the periodic table to determine the mass of a mole of each of the elements and then some. However, many moles of that element is in the compound with all of the other moles of the elements in the compound. So, for example, if we have caesium fluoride, one mole of it, we have one mole cesium and one more chloride. The mass of one mole caesium from the periodic table is 132.91 g per mole. Massive chloride is 35.45 g per mole. We some these up to get the Mueller massive cesium chloride, which is 1 68.36 g per mole for the next one. We have potassium chlorate where in the chemical formula, we have one more potassium, one more chloride, three moles oxygen. So we go to the periodic table and we look up the molar mass of potassium. It's 39.98 We'll add that to the Moller massive chloride 35.45 plus three times the molar mass of oxygen at 15.999 And we'll get a Moeller Mass for the compound of 122.55 g per mole for the next one. We have six moles of carbon at 12.11 g per mole, 12 moles of hydrogen at 1.79 g per mole, and six moles of oxygen at 15.999 g per mole. So we some those up and we get the molar mass of the entire compound, which is 180.155 g per mole for the next one. We have ammonium hydrogen phosphate. This one is a little bit tricky because we have parentheses in our equation. Thes parentheses means we have to pneumonias. That means we have two times one to nitrogen and two times 48 hydrogen. But we actually have another hydrogen here, So in one mole of this compound we have nine moles of hydrogen, so we'll take two times the Mueller massive nitrogen, plus nine times similar massive hydrogen, plus the Mueller massive phosphorus. We just have one of those, plus four times the Mueller Mass of oxygen. I will get 132.55 g per mole for the last one c two h five n 02 We have two moles of carbon plus five moles off hydrogen with the molar mass of 1.79 plus one mole of nitrogen and two moles of oxygen. This gives us a Mueller Mass of the compound of 75.67 g per mole.