Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

# Calculate the number of moles of the indicated substance present in each of the following samples.a. 1.28 $\mathrm{g}$ of iron(II) sulfateb. 5.14 $\mathrm{mg}$ of mercury(II) iodidec. 9.21$\mu \mathrm{g}$ of tin(IV) oxided. 1.26 $\mathrm{lb}$ of cobalt(II) chloridee. 4.25 $\mathrm{g}$ of copper(II) nitrate

## A. $8.43 \times 10^{-3} \mathrm{mol}$B.$1.13 \times 10^{-5} \mathrm{mol}$C.$6.11 \times 10^{-8} \mathrm{mol}$D.4.41 $\mathrm{mol}$E.$2.27 \times 10^{-2} \mathrm{mol}$

Composition

### Discussion

You must be signed in to discuss.
##### Stephanie C.

University of Central Florida

##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto

### Video Transcript

five different compounds, and we need to find a number of moles given a particular months. So if you recall the formula, the number of moles is equivalent to mass divided by more mass. All right, so now we have less. Although we need to find is more mass and to find the more massive a compound. What we do is some up the individual molar masses of each Adam taking into account of the number of each Adam. Right. So I have already gone ahead and how he made it, All of them, all the masses. So this kid get a little bit more quicker. So for here, this first question we have iron saw feet. I hug you with molar mass as being 151.9 grams per mole. Right. So, men, once we do the division, we get 8.4 26 firms tend to the exponents. Negative three. No. And here this e negative three means tend to the exponents. Negative three. All right, Next up, we have mercury. I died 5.45 point 14 times. Tend to the exponents. Negative three grams fit. So the molar mass of mercury I died is 400 and 54.4, Gramps problem. Oops, grams per mole. And once we do this division, we get a number of moles being 1.1 three u times tend to the exponents. Negative five nos. Next up, we have 10 oxide, and here we have micrograms. So we need to do the unit conversion. So the way this works is we multiply if quantity by a quantity, that is equivalent to one where we have micrograms in the denominator to cancel out. And we have the desired unit Gramps in the numerator. So Okay, we know that one gram is equivalent to one things. 10 to the six micrograms. So because it's equivalent this everything in this bracket is equivalent to one. So we're not changing this Quantum E if you run through this too division we yet Let me bring my cut clear out. I'm going to six. Divided by one e six. Oh. Yep. It's 10 9.26 e negative six. I should have known grams, right? So then we take this mass and divide it by the molar acid Oxlade, which is 150 points, seven grams per mole until you I'm going to erase this to now state What this number of moles is equivalent to is basically 150 Poland with all of us, 6.11 most okay, Next we have cobalt chlorate, opens expats. So again we have to do another unit conversion. And it's the exact same ways I've just explained. We need to multiplies. Is quantity by, ah unit conversion that is basically equivalent to one with pounds in the denominator in the desired unit fans in the numerator. Also, it's important for us to, um, come right because we are then going to divide by molar Mass, which is in wheezing Gramps. So it's easier if you just came Brit Mass instead of worrying about converting all of us. Okay, so it's known that one pound is equivalent to 453 cool. 400 and 53 0.592 um, Grams. Yeah, that's the unit conversions. So now that we have grams weaken, divide the Rams by the molar mass of Kobol chloride, which 129 0.83 86 grams per mole. All right, so if we work through this, We yet 4.4 moles. Lastly, we have copper, my tricked and 4.25 grams of it. So the molar mass of copper nitrate is 187 point wife I grams per mole. Kidding? It was equals. Zero Put with 0 to 2 sis. 0.0 2 to 65 This all right?

McMaster University

#### Topics

Composition

##### Stephanie C.

University of Central Florida

##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto