Science, asked by tin3hhh, 17 days ago

Compute the Molecular Formula of the ff:
a. A compound has the E.F. Mo2Cl3Br5with a molar mass of 300 g/mol.
b. A compound has the EF.CH2O with the molar mass of 180.18 g/mol.
c. A compound containing Nitrogen(13.8 g), and Oxygen (4.2 g) only has a molar mass of 90.3 g/mol. Determine the molecular formula of the compound.

Answers

Answered by vihanmaurya11may2012
0

Answer:

Divide the molar mass of the compound by the empirical formula molar mass. The result should be a whole number or very close to a whole number. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula

Explanation:

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Answered by muhammedraedchakarka
0

Answer:

ANSWER IS

Explanation:

Every time a problem provides you with a compound's molar mass, you can use a little trick to help you determine the compound's molecular mass faster.

More specifically, instead of determining the empirical formula first, then using the molar mass to get the molecular formula, you can skip the empirical formula altogether.

As you know, a compound's molar mass tells you what the mass of one mole of that substance is. In this case, one mole of your compound has a mass of 142.00 g.

This means that if you pick a sample of 142.00 g of this compound, you can use its percent composition to get the exact number of moles of each element you get per mole of compound.

So, you know that this compound's percent composition is as follows

carbon→50.7 %

hydrogen→4.2%

oxygen→45.1 %

This means that it contains

142.00 g compound×

100 g compound

50.7 g C

=71.994 g C

142.00 g compound×

100 g compound

4.2 g H

=5.964 g H

142.00 g compound×

100 g compound

45.1 g O

=64.042 g O

Now, all you have to do is use the molar masses of these three elements to determine how many moles of each element you get in one mole of your compound.

For C: 71.994 g×

12.011 g

1 mole C

=5.994≈6

For H: 5.964 g×

1.00794 g

1 mole H

=5.92≈6

For O: 64.042 g×

15.994 g

1 mole O

=4.003≈4

Therefore, the molecular formula of the compound will be C

6

H

6

O

4

.

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