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Rufina [12.5K]
3 years ago
7

What is the percent by mass of oxygen in fe2o3

Chemistry
2 answers:
Effectus [21]3 years ago
5 0
To get the formula mass we have to add the atomic mass of each atom
(55.847 x 2) + ( 16 x 3) = 160
this means that 16 x 3 = 48 are the grams of O
% = 48 x 100 / 160 =30 %

hope this helped mark me brainliest
anygoal [31]3 years ago
4 0
30% is the answer. Good luck
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Thank you for posting your question here. I am just assuming to that the density of tin is <span>7.31 g/cm^3, you can change it it is not the number. Below is the solution. I hope it helps. 

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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di
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<u>Answer:</u> The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M

  • <u>Calculating the molarity of solution:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (camphor) = 70 g

M_{solute} = Molar mass of solute (camphor) = 152.2  g/mol

W_{solvent} = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m

  • <u>Calculating the mole fraction of camphor:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For camphor:</u>

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol

<u>For ethanol:</u>

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

\chi_{(camphor)}=\frac{0.459}{10.272}=0.045\

  • <u>Calculating the mass percent of camphor:</u>

To calculate the mass percentage of camphor in solution, we use the equation:

\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

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O2 is the limiting reactant 

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