All categories

3 years ago

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

Physics

2 answers:

Black_prince [1.1K]3 years ago

6 0

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: brainly.com/question/5950133

2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603

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.

allsm [11]3 years ago

4 0

First, we need to know the amounts of the elements in the compound.

Tin (Sn)= 5.28 g
Fluorine (F) = 8.65 - 5.28 = 3.37 g

Convert these to units of moles by dividing the molar masses.

Tin (Sn)= 5.28 g / 118.71 g/mol = 0.044 mol
Fluorine (F) = 3.37 g / 19.00 g/mol = 0.177 mol

Divide both by the least number of moles of the two.

Tin (Sn)= 0.044 mol / 0.044 mol = 1
Fluorine (F) = 0.177 mol / 0.044 mol = 4

Therefore, the empirical formula would be:
SnF4

You might be interested in

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

PART C) The gravitational force on the satellite is given by,

$F = ma$

$F = \frac{1}{4} mg$

$F = \frac{270*9.8}{4}$

$F = 661.5N$

5 0

3 years ago

A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with

Wittaler [7]

2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!

8 0

2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .