Limiting Reactants?

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Daniela_Chem14A
Posts: 27
Joined: Tue Nov 15, 2016 3:00 am

Limiting Reactants?

Postby Daniela_Chem14A » Mon Jul 10, 2017 2:53 pm

During the review session, Professor Lavelle showed the following example (also on the attachment):

L-Dopa, a drug used for the treatment of Parkinson's disease is 54.82% of C, 5.62% of H, 7.10% of N and 32.46% of O and a molar mass of 197.19g/mol.

My question is why does he multiply the percentage by the molar mass given to convert to grams rather than just assuming that there are 100g being used overall, resulting in 54.82g of C, 5.62g of H, 7.10g of N and 32.46g of O?

Thank you!
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Stephanie H
Posts: 20
Joined: Fri Jun 23, 2017 11:39 am

Re: Limiting Reactants?

Postby Stephanie H » Mon Jul 10, 2017 6:12 pm

When we are not given a molar mass, we can assume there are 100g of the sample. However, since the molar mass was given we can use that to find the actual amounts.

Johana Jeon 1A
Posts: 30
Joined: Sun Feb 12, 2017 3:00 am

Re: Limiting Reactants?

Postby Johana Jeon 1A » Mon Jul 10, 2017 7:01 pm

Stephanie H wrote:When we are not given a molar mass, we can assume there are 100g of the sample. However, since the molar mass was given we can use that to find the actual amounts.



Then how come for F.17 we don't need to do it that way?
The question is: Osmium forms a molecular compound with mass percentage composition 15.89% C, 21. 18% O, and 62.93% Os.
a) What is the empirical formula of this compound?
b) From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g mol^-1 What is its molecular formula?

Thank you in advance

Chels Zh 1D
Posts: 41
Joined: Fri Jun 23, 2017 11:39 am

Re: Limiting Reactants?

Postby Chels Zh 1D » Mon Jul 10, 2017 7:13 pm

For step (a), I would like to go with to assume there are 100g of the sample, I got the ratio is 4:4:1 of C:O:Os. Then I got the empirical formula OsC4O4. Then at (b) we can calculate the molar mass of the empirical formula, which is 302.24 g/mol, divided this number by 907 g/mol, which gives us a new ratio is 3:12:12. Os3C12O12 is the molecular formula.

Stephanie H
Posts: 20
Joined: Fri Jun 23, 2017 11:39 am

Re: Limiting Reactants?

Postby Stephanie H » Mon Jul 10, 2017 7:24 pm

Johana Jeon 1A wrote:
Stephanie H wrote:When we are not given a molar mass, we can assume there are 100g of the sample. However, since the molar mass was given we can use that to find the actual amounts.



Then how come for F.17 we don't need to do it that way?
The question is: Osmium forms a molecular compound with mass percentage composition 15.89% C, 21. 18% O, and 62.93% Os.
a) What is the empirical formula of this compound?
b) From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g mol^-1 What is its molecular formula?

Thank you in advance


We have to treat each question separately, in part a they do not give us a molar mass, so we have to assume there is 100g.

Johana Jeon 1A
Posts: 30
Joined: Sun Feb 12, 2017 3:00 am

Re: Limiting Reactants?

Postby Johana Jeon 1A » Mon Jul 10, 2017 7:27 pm

Stephanie H wrote:
Johana Jeon 1A wrote:
Stephanie H wrote:When we are not given a molar mass, we can assume there are 100g of the sample. However, since the molar mass was given we can use that to find the actual amounts.



Then how come for F.17 we don't need to do it that way?
The question is: Osmium forms a molecular compound with mass percentage composition 15.89% C, 21. 18% O, and 62.93% Os.
a) What is the empirical formula of this compound?
b) From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g mol^-1 What is its molecular formula?

Thank you in advance


We have to treat each question separately, in part a they do not give us a molar mass, so we have to assume there is 100g.


Got it, thank you so much!


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