Chemistry Molecular Formuals from Empirical Formula

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The discussion focuses on determining the empirical and molecular formulas of an unknown compound containing 21.8 g of phosphorus and 28.2 g of oxygen, with a molar mass of 284 g/mol. Initial calculations yield an empirical formula of PO3 based on a mole ratio of approximately 1:3. However, a participant suggests refining the ratio to 5:2, leading to the molecular formulas P2O5 or P4O10, which align with the oxidation states of phosphorus and oxygen. The calculations emphasize the importance of accurate mole ratios in deriving chemical formulas. Ultimately, the correct molecular formula reflects the balance of oxidation numbers between the elements involved.
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Homework Statement


An unknown compound contains 21.8 g of phosphorus and 28.2 g of oxygen. It's molar mass is 284 g/mol. what are its empirical and molecular formulas?

Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol



2. The attempt at a solution

This is my math:
Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

Converting the given measurements to moles:
21.8gP / 31 = .70 mol P
28.2gO / 16 = 1.8 mol O

Getting the ratio:
.70 / .70 = 1
1.8 / .70 = 2.57 (rounds up to 3)

Ratio:
1:3

Empirical Formula : PO3

Formula Mass of PO3:
79g/mol

Finding the common factor:
284(the given weight)/ 79(the weight of PO3) = 3.594 (rounds up to 4)

Writing the Molecular Formula:
4(PO3) = P4O12

Getting the Molar Mass of the Molecular Formula:
P4 = 124
O12 = 192
P4O12 = 316
316 =/= 284(is what it should be)
 
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Nevermind, I did some research and found out that if the ratio is ~2.5 that you're supposed to multiply it by 2 to get even numbers.
 
HigueraC said:

Homework Statement


An unknown compound contains 21.8 g of phosphorus and 28.2 g of oxygen. It's molar mass is 284 g/mol. what are its empirical and molecular formulas?

Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol



2. The attempt at a solution

This is my math:
Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

Converting the given measurements to moles:
21.8gP / 31 = .70 mol P
28.2gO / 16 = 1.8 mol O

Getting the ratio:
.70 / .70 = 1
1.8 / .70 = 2.57 (rounds up to 3)

Ratio:
1:3

Empirical Formula : PO3

Formula Mass of PO3:
79g/mol

Finding the common factor:
284(the given weight)/ 79(the weight of PO3) = 3.594 (rounds up to 4)

Writing the Molecular Formula:
4(PO3) = P4O12

Getting the Molar Mass of the Molecular Formula:
P4 = 124
O12 = 192
P4O12 = 316
316 =/= 284(is what it should be)
====================
You got a ratio of 2.57:1. instead of rounding to 3, try 5:2, which is a lot closer.
Then your formula is P2O5 or P4O10. This makes sense since P and O have oxidation numbers of +5 and -2, so the electrons balance out to zero.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

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