Physics, 11.01.2020 04:31 cam961

If 8.65 g of a tin - fluorine compound contains 5.28 g of tin, what is its empirical formula?

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Answer from: biancatay

The empirical formula of the tin-fluorine compound is $\boxed{{\text{Sn}}{{\text{F}}_{\text{4}}}}$ .

Further explanation:

Empirical formula:

It is atom’s simplest positive integer ratio in the compound. It may or may not be same as that of molecular formula. For example, empirical formula of sulfur dioxide is ${\text{SO}}$ .

Step 1: Mass of fluorine in tin-fluorine compound is to be calculated. This is done by using equation (1).

Since the compound consists of only tin (Sn) and fluorine (F). So the mass of fluorine is calculated as follows:

${\text{Mass of F}} = {\text{Mass of compound}} - {\text{Mass of Sn}}$ …… (1)

The mass of the compound is 8.65 g.

The mass of Sn is 5.28 g.

Substitute these values in equation (1).

$\begin{aligned}{\text{Mass of F}}&={\text{8}}{\text{.65 g}}-{\text{5}}{\text{.28 g}}\\&={\text{3}}{\text{.37 g}}\\\end{aligned}$

Step 2: The moles of tin and fluorine are to be calculated.

The formula to calculate the moles of a substance is as follows:

${\text{Moles of substance}} = \frac{{{\text{Given mass of substance}}}}{{{\text{Molar mass of substance}}}}$ …… (2)

Substitute 5.28 g for given mass and 118.71 g/mol for molar mass in equation (2) to calculate the moles of Sn.

$\begin{aligned}{\text{Moles of Sn}}&=\left({{\text{5}}{\text{.28 g}}}\right)\left( {\frac{{{\text{1 mol}}}}{{{\text{118}}{\text{.71 g}}}}}\right)\\&=0.044{\text{4 mol}}\\\end{aligned}$

Substitute 3.37 g for given mass and 18.99 g/mol for molar mass in equation (2) to calculate the moles of F.

$\begin{aligned}{\text{Moles of F}}&=\left({{\text{3}}{\text{.37 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{18}}{\text{.99 g}}}}}\right)\\&=0.1774{\text{ mol}}\\\end{aligned}$

Step 4: The preliminary formula of tin-fluorine compound is to be formed.

The moles of tin and fluorine are to be written with their corresponding subscripts in nitrogen oxide. So the preliminary formula becomes,

${\text{Preliminary formula of tin - fluorine compound}} = {\text{S}}{{\text{n}}_{0.0444}}{{\text{F}}_{0.1774}}$

Step 5: The empirical formula of tin-fluorine compound is to be formed.

Each of the subscripts is divided by the smallest subscript to get the empirical formula. In this case, the smallest one is 0.0444. So the empirical formula of the compound is written as follows:

$\begin{aligned}{\text{Empirical formula of tin - fluorine compound}}&={\text{S}}{{\text{n}}_{\frac{{0.0444}}{{0.0444}}}}{{\text{F}}_{\frac{{0.1774}}{{0.0444}}}}\\&={\text{S}}{{\text{n}}_{\text{1}}}{{\text{F}}_{{\text{3}}{\text{.99}}}}\\&\approx{\text{Sn}}{{\text{F}}_{\text{4}}}\\\end{aligned}$

So the empirical formula of the compound comes out to be ${\mathbf{Sn}}{{\mathbf{F}}_{\mathbf{4}}}$ .

Learn more:

1. Calculate the moles of ions in the solution:

2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stoichiometry of formulas and equations

Keywords: empirical formula, Sn, F, SnF4, tin, fluorine, compound, subscript, moles of tin, moles of oxygen, mass of fluorine, mass of tin, moles of fluorine, preliminary formula.

Answer from: Quest

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

Answer from: Quest

answer; equivalent to a one(1) horsepower motor;

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