The empirical formula of an oxide of carbon which has 52.94 percent carbon and 47.06 percent oxygen by mass is
Answers
Answer:
i dont know how to solve youe example but i know the correct definition of empirical formula
Explanation:
The empirical formula for a compound is one that expresses the smallest whole number ratio of atoms present. What this means is that the empirical formula is merely a ratio of the different elements that comprise that compound. A wonderful example is water, H2O. This is the empirical formula. For every two hydrogen atoms there will be one oxygen atom. If there were 6 hydrogen atoms and 3 oxygen atoms, the formula would be H6O3. You would still reduce this down to the ratio of H2O. Leaving it as H6O3 (which is merely an example and not any true compound that I know) would be the molecular formula – which we’ll get to in due time.
So how do you calculate empirical formulas? Well, there are a defined set of steps to follow:
Determine the mass in grams of each element present. (If you are given percentages in a problem, then assume the sample has 100g).
Calculate the number of moles of each element.
Divide each by the smallest number of moles to obtain the simplest whole number ratio.
If step 3 does not yield whole numbers, then try multiplying by any factor to get whole numbers.
The only other disclaimer I can give you for these problems is to be sure to make accurate calculations. Rounding error can completely throw off your end ratio. After all, 14.50067 is way more accurate than 14.5!
Example: A sample contains 2.34 g of Nitrogen and 5.34 grams of Oxygen. What is the empirical formula?
The first step is to determine the mass in grams. This has already been given in the problem.
Now we must calculate the number of moles. (If you have trouble with this, please see my previous explanatory blog on how to do so). We must do this both for Nitrogen and Oxygen.
2.34 grams x (1 mole / 14.007 g N) = 0.167 moles Nitrogen.
2.34 grams x (1 mole / 15.999 g O) = 0.333 moles of Oxygen.
Now, to get a whole number ratio, we must divide each by the smaller of the two numbers. This will be the 0.167 moles of Nitrogen.
0.167 / 0.167 = 1.
0.333 / 0.167 = 1.99 = 2.
The end result is one Nitrogen atom for every two Oxygen atoms. So the empirical formula is NO2. How about one more, just to be sure?
Example: The chemical analysis of a compound indicates that a compound has 52.94% aluminum and 47.06% oxygen. Determine the empirical formula for this compound.
The first step is to determine mass in grams. Remember that, if given percentages, to assume a sample size of a 100g. That would mean 52.94 grams of Aluminum, and 47.06 grams of Oxygen.
Now we need to calculate the number of moles present. This would be done like so:
Aluminum: 52.94 g x (1 mole / 26.482 g) = 1.96 moles Al.
Oxygen: 47.06 g x (1 mole / 16 g) = 2.94 moles O.
Now, as before, we need to divide by the smallest number. This would be Aluminum’s 1.96.
1.96 / 1.96 = 1.
2.94 / 1.96 = 1.5
Our current ration is one Aluminum for every 1.5 Oxygen. This isn’t necessarily a “clean” answer. So, in this case it is permissible to multiply everything by a common number to make a better overall answer. In this case, multiply both quantities by two. This gives: Al2O3. Notice how the ration is still the same.