# Empirical formula when given percentages?

A binary compound of lead and oxygen contains $90.66 \% \ce{Pb}$. What is the empirical formula for the compound? How do I use the percentage to get to the normal formula, then how do I change the normal formula to empirical?

• In these kinds of questions (homework), we like to see what you have tried on this problem so that we can provide a more focused answer. Without knowing where you are stuck, you are likely to get an answer very similar to the example problem in your textbook or your instructor's advice (which obviously weren't helpful or you wouldn't have posted here). Commented Apr 28, 2013 at 10:31
• Ah sorry actually I had no idea of how to even begin the question! Commented Apr 29, 2013 at 9:10

You will get the empirical formula straight from this data. In order to convert to a molecular formula, you need more information.

By way of another example:

A binary compound of tin and chlorine contains 62.89% tin by mass. Determine the empirical formula.

Our first goal is to figure out the ratio of moles of tin to chlorine in this compound. Once we know the ratio of moles, we can determine the formula.

What do we know? We know what percentage of tin is in this compound by mass. What is the percentage of chlorine?

In order to get to moles, we need at least a mass of the sample. We were not given that. Since percent by mass is a ratio of masses, it turns out we can choose any mass of sample we want and get the same formula (Try it yourself!).

If we assume that we started with 100.0 g of this stuff, then how many grams of tin would you have? How many grams of chlorine would you have? 100.0 g is a nice starting mass since you are working with percents.

Now, as Ari's answer says, you need to convert mass of tin and mass of chlorine into moles of tin and moles of chlorine by dividing each mass by the atomic molar mass of each element (look it up on your periodic table).

Now that you are in moles, determine the lowest whole number ratio of the two elements by dividing each number of moles by the smaller number of moles. If your numbers are close to whole numbers, you can round. For example, round 1.97 to 2. If your numbers are not close to whole numbers, you may need to do a little more arithmetic:

• If you get a number like 1.48, this is close to 1.5 which is 3/2. Multiply both numbers by 3.
• If you get numbers like 1.34 (close to 4/3) or 1.67 (close to 5/3) multiply by 3.

Now you have the empirical formula. As Ari's answer says, you need a molar mass of the species to get a molecular formula.

The answer for my example is $\ce{SnCl2}$. Work through the steps and see if you can get to this answer. Then repeat for your lead and oxygen compound.

• Can I ask one more thing, I tried it I have the answer for your example and my example but for my question it says the answer is Pb3O4 where as I worked it out to be one : one moles in ratio, so PbO, the answers state that they are three : four atom ratio could you explain that by anychance? Commented Apr 30, 2013 at 10:27
• When I went through the maths, I got $\ce{Pb3O4}$. Moles of lead $=90.66/207.2=0.4375$. Moles of oxygen $=9.34/16.00=0.5838$. Divide by the smaller number $0.4375$, and you get $\ce{Pb1O}_{1.33}$. 1.33 is too far from one to round (but it is $\frac{4}{3}$ to three sig figs). Multiply by 3: $\ce{Pb}_{1\times3} \ce{O}_{1.33\times3}=\ce{Pb3O4}$. Commented Apr 30, 2013 at 10:47
• I see ^^ thankyou soo much! your explanation is so clear cut, thankyou for helping me to understand these questions! keep up the great work :) Commented Apr 30, 2013 at 11:16

I'm not going to this question for you. The point of this forum is to educate and not to do your homework.

Imagine you 100 g of compound XY where X and Y are elements.

You know the percentage of X is 70 and of Y is 30.

This means there must be 70 g and 30 g of X and Y respectively.

To convert this to moles you need to do :

$\frac{70}{molar\;mass\;of\;X}$ and similarly for Y.

Then make use of the definition of the empirical formula.

NOTE : To obtain the molecular formula you need the molecular mass.

• Just letting you know I did attempt this question before asking and Im not using this site as a homework help, Im genuinely trying to find out how to do empirical formula, and this actually isnt homework its extra questions I am doing doing myself to prepare and brush up on my chem just clearing that up. Commented Apr 29, 2013 at 9:09