6.10g of boron reacted completely with 22.9g of chlorine to give 29.0g of the metallic chloride. Calculate the empirical formula of the chloride.
I've gone through this time and time again and something isn't clicking. Here is the way to do it BUT the answers are not reasonable.
%B = (6.1/29)*100 = 21.03
%Cl = (22.9/29)*100 = 78.97
Take 100 g sample which then gives you 21.03 g B and 78.97 g Cl.
mols B = 21.03/10.81 = 1.95
mols Cl = 78.97/35.45 = 2.23
B = 1.95/1.95 = 1
Cl = 2.23/1.95 = 1.14
Then you multiply both values by a whole number; i.e. by 2, 3, 4, 5, 6, etc until you come up with numbers that can be rounded to whole numbers. Multiplying by 7 gives 1*7 = 7 for B
1.14*7 = 7.98 which rounds to 8.0 and that gives B7Cl8. As far as I know there is no such compound. The person making up the problem either made a typo, didn't calculate it right, OR just made up numbers that resulted in no such compound. Some profs do that but I never did. I think it's unfair to make up fake numbers to give fake compounds.
I AGRÉE WITH YOU
agreed
That's true, thanks
Use the following method:
Mass
Formula Mass
Moles
Divide by smallest ratio
Ratio
I like the explicandum, it is very understandable.
Coming to the last part, its not just 7 that can round it up. Check again. Numbers b4 7. Just My observation
I know there is only BCl3
B8Cl8
B9Cl9
Let's look into these instead
To determine the empirical formula of the chloride, we need to find the relative number of atoms of each element present in the compound.
First, let's find the number of moles of each element using their respective molar masses:
- Moles of Boron (B) = Mass of Boron / Molar Mass of Boron
= 6.10g / 10.81g/mol (molar mass of B)
≈ 0.563 mol
- Moles of Chlorine (Cl) = Mass of Chlorine / Molar Mass of Chlorine
= 22.9g / 35.45g/mol (molar mass of Cl)
≈ 0.647 mol
Next, we need to determine the molar ratio between Boron and Chlorine atoms. We can achieve this by dividing the moles of each element by the smaller mole value:
- Moles of B / smaller mole value = 0.563 mol / 0.563 mol (smaller mole value)
= 1
- Moles of Cl / smaller mole value = 0.647 mol / 0.563 mol
≈ 1.15
The empirical formula of the chloride compound can be written as BCl₂, indicating that there is one Boron atom and two Chlorine atoms present.
Therefore, the empirical formula of the chloride compound is BCl₂.