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hram777 [196]
3 years ago
8

The empirical formula of a compound is CH2O and its molecular weight is 90.09 g/mol. Determine the molecular formula.

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer : The molecular formula of the compound will be, C_{3}H_{6}O_3

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.  

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

As we are given that the empirical formula of a compound is CH_2O and the molar mass of compound is, 90.09 gram/mol.

The empirical mass of CH_2O = 1(12) + 2(1) + 1(16) = 30 g/eq

n=\frac{\text{molecular mass}}{\text{empirical mass}}

n=\frac{90.09}{30}=3

Molecular formula = (CH_2O)_n=(CH_2O)_3=C_{3}H_{6}O_3

Thus, the molecular formula of the compound will be, C_{3}H_{6}O_3

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3 years ago
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
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Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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