Question

Oxides of virtually every element are known. bromine, for example, forms several oxides when treated with ozone. suppose you allow 1.250 g of bromine, br2, to react with ozone, o3, and obtain 1.876 g of brxoy. what is the empirical formula of the product?

Answer 1

Answer is: empirical formula of the product is Br₂O₅.
Chemical reaction: x/2Br₂ + y/3O₃ → BrₓOy.
m(Br₂) = 1,250 g.
m(BrₓOy) = 1,876 g.
n(Br₂) = m(Br₂) ÷ M(Br₂).
n(Br₂) = 1,25 g ÷ 159,81 g/mol.
n(Br₂) = 0,0078 mol.
n(Br) = 2 · 0,0078 mol = 0,0156 mol.
m(O₃) = 1,876 g - 1,25 g = 0,626 g.
n(O₃) = 0,626 g ÷ 48 g/mol = 0,013 mol.
n(O) = 0,039 · 3 = 0,039 mol
n(Br) : n(O) = 0,0156 mol : 0,039 mol.
n(Br) : n(O) = 1 : 2,5.

Answer 2

Answer:

$Br_2O_5$

Explanation:

Hello,

Considering the given information, the undergoing chemical reaction is:

$Br_2+O_3-->Br_xO_y$

Thus, we consider the formed grams of $Br_xO_y$ because the grams of bromine are equal before and after the chemical reaction (mass can't be neither created nor destroyed), thus, the bromine grams into the $Br_xO_y$ are 1.250g and the oxygen grams that come from the ozone are:

$m_O=1.876gBr_xO_y-1.250gO=0.626gO$

Now, we compute the moles for each one of them as:

$n_{Br}=1.250gBr_2*\frac{1molBr_2}{160gBr_2}*\frac{2molBr}{1molBr_2}=0.0156molBr.\\ n_O=0.626gO*\frac{1molO}{16gO}=0.039molO$

Now, we divide by the bromine's moles to find the littlest whole-number that allows us to identify the empirical formula as shown below:

$Br=\frac{0.0156}{0.0156}=1;O=\frac{0.039}{0.0156} =2.5$

Finally, by multiplying by two to find the littlest whole-number, one gets:

$Br_2O_5$

Best regards.