Amineral sample is obtained from a region of the country that has high arsenic contamination. an elemental analysis yields the following elemental composition:

element atomic weight (g/mol) percent composition
ca 40.078 22.3%
as 74.9216 41.6%
o 15.9994 35.6%
h 1.00794 60%

what is the empirical formula of this mineral?

CaAsHO₄

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

Step 1. Calculate the amount in moles of each element.

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

Step 3. Write the empirical formula with the rounded numbers obtained

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

CaAsHO₄

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.

i dont know

explanation: