Question

Question #c7a58

Answer 1

The total mass of the sample is `5.0` `g` but of that `1.51` `g` is potassium (`K`), which leaves `5.00-1.51=3.49` `g` which is comprised of `Br` and `I`.

The molar mass of potassium is `39` `gmol^-1`, so we can find the number of moles of `K`:

`n=m/M=1.51/39=0.039` `mol`

Since all the `Br` and `I` are present as `KBr` and `KI`, we know that the total number of moles of `Br+I=0.039` `mol`.

The molar masses are:

`Br - 79.9` `gmol^-1`

`I - 126.9` `gmol^-1`

Call the number of moles of `Br` 'n_(Br)', then the number of moles of `I` will be `n_I=0.039-n_(Br)`.

Similarly, if we call the mass of `Br` 'm_(Br)', then the mass of `I` will be `M_I=3.49-m_(Br)`.

We have:

`n_(Br)=m_(Br)/M_(Br)`

`n_I=(0.039-n_(Br))=m_I/M_I=m_I/(3.49-M_(Br))`

I'm stuck here for the moment. Will try to solve it in the morning, or someone else can fix it instead. Sorry: just to sleepy to make sense of it.

Answer 2

`"KBr = 71.4 %; KI = 28.6 %"`

Explanation:

General method

Usually, we solve this problem by setting up two simultaneous equations in two unknowns.

Here, the two equations would be

`bb"(1)"color(white)(m) "Mass of KBr + Mass of KI = Total mass"`

`bb"(2)"color(white)(m) "Mass of K from KBr + Mass of K from KI = Mass of K"`

Step 1. Set up Equation (1)

Let `x = "mass of KBr"` and `y = "mass of KI"`

This gives

`bb"(1)"color(white)(m)x + y = 5.00`

Step 2. Set up Equation (2)

`"Fraction of K in KBr" = 39.10/119.00`

`"Fraction of K in KI" = 39.10/166.00`

This gives

`39.10/119.00x + 39.10/166.00y = 1.51`

`bb"(2)"color(white)(m)0.3286x + 0.2355y = 1.51`

Step 3. Solve the system of equations

From (1),

`bb"(3)"color(white)(m)y = 5.00 - x`

Substitute (3) in (2).

`0.3286x + 0.2355(5.00-x) = 1.51`

`0.3286x + 1.1777 - 0.2355x = 1.51`

`0.0930x = 0.3323`

`x= 0.3323/0.0930 = 3.57`

`y = 5.00 - 3.57 = 1.43`

Step 4. Calculate the percent composition

`"% KBr" = "Mass of KBr"/"Total mass" × 100 % =(3.57 color(red)(cancel(color(black)("g"))))/(5.00 color(red)(cancel(color(black)("g")))) × 100 % = 71.4 %`

`"% KBr" = "Mass of KBr"/"Total mass" × 100 % =(1.43 color(red)(cancel(color(black)("g"))))/(5.00 color(red)(cancel(color(black)("g")))) × 100 % = 28.6 %`

Check:

`"Mass of K" =39.10/119.00 × "3.57 g" + 39.10/166.00 × "1.43 g" = "1.17 g + 0.336 g" = "1.51 g"`

It checks!