E.15  [ENDORSED]

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Michael Lee 2I
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Joined: Fri Sep 29, 2017 7:05 am

E.15

Postby Michael Lee 2I » Thu Oct 05, 2017 3:47 pm

The molar mass of the metal hydroxide M(OH)2 is 74.10 g.mol^-1. What is the molar mass of the sulfide of this metal?

I was confused on finding the molar mass of the sulfide. Is there another number I am supposed to know to use to solve this problem?

Charlotte Shieh 1F
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Joined: Fri Sep 29, 2017 7:04 am

Re: E.15

Postby Charlotte Shieh 1F » Thu Oct 05, 2017 4:34 pm

you find what the metal M is by subtracting the molar mass of (OH)2 from 74.10. you will find that M is calcium, so you just have to calculate the molar mass of CaS (which is the sulfide of the metal)

Kourtney Nham 1L
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Re: E.15

Postby Kourtney Nham 1L » Thu Oct 05, 2017 6:29 pm

Going off what Charlotte said, it helps to look at the problem like 'M' is a variable!

Rachel Lu_dis1H
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Re: E.15

Postby Rachel Lu_dis1H » Thu Oct 05, 2017 8:06 pm

when they say that the molar mass of M(OH)2 is 74.10gmol-1 they're giving you two known elements O and H and one unknown M. To find what M is you subtract the molar mass of M(OH)2 by the molar mass of (OH)2. What you're left is the molar mass of M. You can then match that number with an element on the periodic table with that molar mass (Ca). Now that you know the unknown M which is Ca, you can find the molar mass of CaS which is your answer.

James Nguyen 3G
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Joined: Sat Jul 22, 2017 3:00 am

Re: E.15

Postby James Nguyen 3G » Thu Oct 05, 2017 8:24 pm

Are we expected to understand this concept for the upcoming test? As in sulfides, chlorides, etc.

DAllaf
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Joined: Fri Sep 29, 2017 7:04 am

Re: E.15

Postby DAllaf » Thu Oct 05, 2017 10:54 pm

Hey Michael,

You should begin by treating 'M' as an unknown variable. Then find the molar mass of (OH)2. Oxygen's molar mass is 15.99 g.mol^-1 and Hydrogen's molar mass is 1.008 g.mol^1. So multiply each by two since it has a subscript of two and then add them together to get a total molar mass of 34.00 g.mol^-1. Subtract this total from the given total molar mass of M(OH)2 which was 74.10 g.mol^-1. 74.10 - 34.00 = 40.1. Then by looking at the periodic table we see that Calcium has a molar mass of 40.08 g.mol^-1 which matches our remaining molar mass for variable 'M'. Thus we know 'M' is Calcium. In order to find the sulfide of Calcium you have to add Sulfur. So now you have CaS and the molar mass of this molecule is 72.14 g.mol^-1.

Hope this helps

nelms6678
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Re: E.15

Postby nelms6678 » Sat Oct 07, 2017 9:13 pm

An example from section "E"
What mass of NaHSO4 should you weigh out to obtain .20 mol NaHSO4?

How would you start this problem off? Im confused would this be Mass Percentage Comp.?

Anna Lapuos 3C
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Re: E.15  [ENDORSED]

Postby Anna Lapuos 3C » Sun Oct 08, 2017 12:31 am

You use, m = nM.

The molar mass of NaHSO4 is 120.06 and you plug this value in for M, thus, the equation would look like this:

m = (0.20 mol)(120.06 g mol-1) = 24.01 g

NatalieSDis1A
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Re: E.15

Postby NatalieSDis1A » Sun Apr 08, 2018 11:00 am

This may be a bit obvious but I did not realize that M is not an element. It just stands for metal. I was confused about why people were saying to treat it as a variable but it is literally nothing more than a variable. And a sulfide is a compound with sulfur so it makes sense to combine the molar mass of the element we find and the molar mass of sulfur. Super obvious but it took me a couple minutes to realize this.

alexagreco1A
Posts: 34
Joined: Fri Apr 06, 2018 11:03 am

Re: E.15

Postby alexagreco1A » Sun Apr 08, 2018 6:01 pm

I had this question as well. I understand the explanation that 'M' is the variable, but would we have to know how to go about determining the sulfides for the test this week?

Jada Larson 1F
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

Re: E.15

Postby Jada Larson 1F » Tue Apr 10, 2018 6:13 pm

Is "M" the most commonly used variable for an unknown element? Or are other letters used as variables? Thanks.


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