0
$\begingroup$

Need some help with second part.....can't get it to be balanced and therefore can't find the formula and name! Thanks in advance!

What is the empirical formula of each of the following compounds?

a) A liquid containing 2.0g of hydrogen, 32.1g of sulfur and 64.0g of oxygen.

•   Hydrogen = 2.0g / 1.0g = 2 / 1 = 2
•   Sulfur = 32.1g / 32.1g = 1 / 1 = 1
•   Oxygen = 64.0g / 16.0g = 4 / 1 = 4

Empirical formula of this compound is: $\ce{SH_2O_4}$ and the name of the compound is Sulfuric Acid.

b) A white solid containing 4.0g of calcium, 3.2g of oxygen and 0.2g of hydrogen.

****•   Calcium = 4.0g / 40.1g = 0.09975062344 = 
•   Oxygen = 3.2g / 16g = 0.2 = 
•   Hydrogen = 0.2g / 1g = 0.2 =**** 

Empirical formula of this compound is:

c) A white solid containing 0.243g of magnesium and 0.710g of chlorine

•   Magnesium = 0.243g / 24.3g = 0.01 / 0.01 = 1
•   Chlorine = 0.710g / 35.5g = 0.02 / 0.01 = 2

Empirical formula of this compound is: $\ce{MgCl_2}$ and the name of the compound is Magnesium Chloride.

$\endgroup$
  • 1
    $\begingroup$ You know about rounding and significant figures, right? $\endgroup$ – Ivan Neretin Jul 4 '16 at 14:35
  • $\begingroup$ For the first question, it should $\ce{H_2SO_4}$ because usually when we write acid we always start with hydrogen first $\endgroup$ – Simon-Nail-It Jul 6 '16 at 6:10
2
$\begingroup$

You find #2 difficult because #1 and #3 are ridiculously easy!

In these two questions the mass of every element present just happens to match exactly with the atomic weight of that element. So when you divide each mass by the corresponding atomic weight, the resulting ratios are very close to integers, or integers divided by 100. So the (almost) final step of dividing each ratio by the smallest ratio goes by almost unnoticed.

But consider for the moment what would happen if you used three or four decimal place atomic weights. All those nice neat exact answers would disappear in a haze of "sort of" whole numbers, and you would have to use some judgement as to whether a number is close enough to a whole number.

In #2, the questioner made the "drastic" change of using 4.0 and 40.1 as the two numbers for an element. If you had carried on mechanically, dividing all ratios by the smallest ratio, you would have gotten:

  • Ca ... $1.0$
  • O .....$2.005$
  • H......$2.005$

Is $2.005$ close enough to a whole number? Or should you multiply everything by $1000$ to give $1000, 2005, 2005$ as the empirical formula?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.