Mass Percentage Question

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derinceltik1K
Posts: 51
Joined: Sat Aug 17, 2019 12:15 am

Mass Percentage Question

Postby derinceltik1K » Thu Oct 10, 2019 9:51 pm

What is the mass percent of the elements in Sr(NO2)2 ?

how would you start to solve this question?

Deepika Reddy 1A
Posts: 125
Joined: Thu Jul 11, 2019 12:15 am

Re: Mass Percentage Question

Postby Deepika Reddy 1A » Thu Oct 10, 2019 10:16 pm

You would use the molar mass. So find the molar mass of the entire compound, and then the molar mass of all the atoms of each element in the compound. Then divide the molar masses of all the atoms of each element by the total molar mass to get each element's mass percentage composition.

Eugene Chung 3F
Posts: 142
Joined: Wed Nov 15, 2017 3:03 am

Re: Mass Percentage Question

Postby Eugene Chung 3F » Thu Oct 10, 2019 10:21 pm

add the molar masses (the mass shown in the periodic table) for each element in the compound. than, divide each element by the total molar mass of the compound.

Nicholas_Gladkov_2J
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Joined: Sat Aug 17, 2019 12:17 am

Re: Mass Percentage Question

Postby Nicholas_Gladkov_2J » Thu Oct 10, 2019 10:36 pm

derinceltik1K wrote:What is the mass percent of the elements in Sr(NO2)2 ?

how would you start to solve this question?


Percent composition by mass is basically asking what is the percentage of the various elements in a molecule. For your provided example, you would:
1) Find the molar mass of the molecule
2) Find the weight of each element in the compound (atomic mass of each element multiplied by the amount/coefficient/subscript of each element: 1 Sn, 2 N, 4 O)
3) Put each weight of each element over the molar mass, and multiply by 100%

Ayushi2011
Posts: 101
Joined: Wed Feb 27, 2019 12:17 am

Re: Mass Percentage Question

Postby Ayushi2011 » Thu Oct 10, 2019 11:16 pm

I'd start by first finding the how many elements are there in the compound, and then find the molecular mass of the compound. Since we know the number of elements, we divide the molecular mass of each element by the total mass of the compound we just found, to get mass percentage.

salvadorramos3k
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Joined: Wed Sep 18, 2019 12:18 am

Re: Mass Percentage Question

Postby salvadorramos3k » Fri Oct 11, 2019 12:40 am

To find the percent compositions of each element, you would first need to find the molar mass of Sr(NO2)2. Then, you would find the molar masses of each individual element and divide those molar masses by the molar mass of the compound. Then you multiple the decimal by 100%.

AveryAgosto
Posts: 76
Joined: Wed Sep 18, 2019 12:16 am

Re: Mass Percentage Question

Postby AveryAgosto » Fri Oct 11, 2019 11:32 am

To determine mass percentage composition you find the molar mass of the compound then find the molar mass of each element in the compound and divide each elements mass by the total mass and multiply by 100 to find the percentage.

KMenjivar_3A
Posts: 25
Joined: Mon Jan 07, 2019 8:05 am

Re: Mass Percentage Question

Postby KMenjivar_3A » Fri Oct 11, 2019 12:22 pm

To determine the mass percentage, find the total molar mass of the compound. Then, divide it by the molar mass of each element. Lastly, you will need to multiply it by 100, which will give you the mass percentage.

Julianna Laurentano 4G
Posts: 52
Joined: Wed Sep 18, 2019 12:21 am

Re: Mass Percentage Question

Postby Julianna Laurentano 4G » Fri Oct 11, 2019 12:29 pm

Begin by finding the total molar mass of Sr(NO2)2. Then, find the mass of each element and divide that by the total molar mass of Sr(NO2)2 to obtain each percentage. Make sure that your calculations include 1 Sr atom, 2 N atoms, and 4 O atoms.

Catherine Daye 1L
Posts: 104
Joined: Wed Sep 11, 2019 12:16 am

Re: Mass Percentage Question

Postby Catherine Daye 1L » Sat Oct 12, 2019 5:10 pm

First, write down the molar masses of each element in the compound. Then find the molar mass of the compound by adding the atomic masses of the elements (making sure to account for the additional atoms ex. O2 is 32 g/mol). Finally, find the mass percent for each element using the equation [(molar mass)/(mass of compound)]*100.


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