H.25

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Abigail Carter 4G
Posts: 50
Joined: Sat Sep 07, 2019 12:19 am

H.25

Postby Abigail Carter 4G » Sun Oct 06, 2019 11:30 pm

Phosphorus and oxygen react to form two different phosphorus oxides. The mass percentage of phosphorus in one of these
oxides is 43.64%; in the other, it is 56.34%. (a) Write the empirical formula of each phosphorus oxide. (b) The molar mass of the former oxide is 283.33 g.mol and that of the latter is 219.88 g.mol Determine the molecular formula and name of each oxide. (c)Write a balanced chemical equation for the formation of each oxide.

Can someone walk me through the steps on how to go about starting this problem?

ishaa Diwakar 4E
Posts: 56
Joined: Wed Sep 11, 2019 12:17 am

Re: H.25

Postby ishaa Diwakar 4E » Sun Oct 06, 2019 11:39 pm

Since you have the molar masses and the percentage composition, you just convert the percent composition to grams and divide by the molar mass. Then, using the molecular formulas, you can write a chemical reaction and balance it.

Sofia Barker 2C
Posts: 101
Joined: Wed Sep 18, 2019 12:21 am

Re: H.25

Postby Sofia Barker 2C » Wed Oct 09, 2019 10:40 am

43.64% of the first compound is phosphorus so the other 56.36% must be oxygen.
Assuming 100-gram sample, divide 43.64g phosphorus by its molar mass (30.974 g/mol) to get 1.409 mol. The mol of oxygen would be 56.36g / 16 g/mol = 3.523 mol.
You then divide each of these moles by the smallest mole value, which in this case is 1.409, to find the ratio of atoms in the compound.
1.409 / 1.409 = 1 P
3.523 / 1.409 = 2.5 O
So the compound has a ratio of 2 P to 5 O, giving an empirical formula of P2O5.
To find the molecular formula, compare the given molar mass of the sample (283.33 g/mol) to the molar mass of the empirical formula (141.948 g/mol). The quotient of 283.33 g/mol / 141.948 g/mol is around 2, which means that the molecular formula is double that of the empirical formula. Thus the molecular formula is P4O10. Knowing this, you can create a balanced chemical equation in which O and P react to form P4O10.
Use this technique to solve for the other unknown phosphorus oxide.

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

Re: H.25

Postby Chris Tai 1B » Wed Oct 09, 2019 3:32 pm

Solving for the other phosphorus oxide, using a similar method:

First find the percentage of oxygen present in the oxide; since 56.34% of the compound is phosphorus, 43.66% of the compound must be oxygen.

If we pretend that we have 100g of this substance, then we know that 56.34 g would be phosphorus, and 43.66 g would be oxygen based on these percentages.

Next, we can divide both 56.34 g P and 43.66 g O by their respective molar masses, like this:
56.34 g / 30.9 g.mol^-1 = 1.8233 mol P
43.66 g / 16 g.mol^-1 = 2.72875 mol O

Next, divide by the smallest amount of moles to get the respective empirical formula for the oxide by finding the ratio between the two atoms in the compound:
1.8233/1.8233 = 1
2.72875/1.8233 = 1.496

And multiply by 2 to get two roughly whole numbers 2 and 3. Hence, the empirical formula is P2O3. The molar mass of this substance is (2)(30.9)+(3)(16) = 109.8 g.^-1
Since 219.88g.mol^-1 / 109.8 g^-1 is roughly 2, the molecular formula is P4O6.


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