How do I do the empirical formula for 37.01% carbon, 2.22% hydrogen, 18.50% nitrogen, and 42.27% oxygen? My teacher gave me the answer (C7H5N3O6), but I can never get that answer.

Take a 100 g sample which gives you

37.01g C]
2.22 g H
18.50 g N
42.27 g O

Convert grams to mols mols = grams/atomic mass.

37.01/12 = approx 3.08
2.22/1 = approx 2.22
18.50/14 = approx 1.32
42.27/16 = approx 2.64

Now find the ratio of these elements to each other with the smallest number no smaller than 1.00. The easy way to do that is to divide the smallest number by itself, then divide all of the other numbers by the same small number.
3.08/1.32 = approx 2.33
2.22/1.32 = 1.68
1.32/1.32 = 1.00
2.64/1.32 = 2.00

You want whole numbers but 0.33 and 0.68 are too large to be thrown away (and if you do this more accurately than my approximations above it may be a little better but you still must try and find whole numbers.) What I do is multiply all of the numbers by 2,3,4,5 etc until you get whole numbers (or at least very close to whole numbers). I can see that the way to get rid of that .33 is to multiply by 3 so let's try that. (Try multiplying by 2 and see what you get.)
3.08/1.32 = approx 2.33*3 = 6.99 C
2.22/1.32 = 1.68*3 = 5.04 H
1.32/1.32 = 1.00*3 = 3.00 N
2.64/1.32 = 2.00*3 = 6.00 O
6.99 will round to 7.0 easily for C.
5.04 rounds to 5.0 for H.
and 3.00 and 6.00 are ok so the empirical formula is C7H5N3O6

Well, don't worry, you're not alone in feeling stumped by empirical formula calculations! It can be a bit confusing at times. But fear not, my friend, I'm here to bring a smile to your face and hopefully shed some light on the matter.

To determine the empirical formula, you need to find the simplest ratio of atoms in a compound. So let's break it down step by step, shall we?

First, you'll want to determine the moles of each element present in the compound. You can do this by assuming you have 100 grams of the compound, as percentages given usually represent grams.

Next, divide the grams of each element by their respective atomic masses to get the moles. For example, for carbon, you'll divide 37.01 grams by the atomic mass of carbon.

Once you have the moles of each element, divide them by the smallest number of moles calculated. This step brings you to the simplest ratio of atoms.

Finally, round off any decimal places to get whole numbers while still maintaining the ratio.

So, give it another whirl, keeping in mind these steps, and see if you can land on the same answer as your teacher. And remember, even if the formulas seem puzzling at times, don't let them rain on your parade! Keep trying, and you'll get there!

To find the empirical formula, we need to determine the simplest ratio of atoms in the compound based on the given percentages. Here's how you can calculate the empirical formula step-by-step:

Step 1: Assume you have a 100 g sample of the compound. This assumption allows the percentages to represent grams, making the math easier.

Step 2: Convert the given percentages to grams by multiplying each percentage by the assumed 100 g. This will give you the grams of each element present in the sample.

37.01% carbon = 37.01 g carbon
2.22% hydrogen = 2.22 g hydrogen
18.50% nitrogen = 18.50 g nitrogen
42.27% oxygen = 42.27 g oxygen

Step 3: Convert the grams of each element to moles. To do this, divide the grams of each element by its molar mass. The molar masses are as follows:
Carbon (C) = 12.01 g/mol
Hydrogen (H) = 1.01 g/mol
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol

Carbon:
moles of carbon = grams of carbon / molar mass of carbon
moles of carbon = 37.01 g / 12.01 g/mol ≈ 3.08 mol

Hydrogen:
moles of hydrogen = grams of hydrogen / molar mass of hydrogen
moles of hydrogen = 2.22 g / 1.01 g/mol ≈ 2.20 mol

Nitrogen:
moles of nitrogen = grams of nitrogen / molar mass of nitrogen
moles of nitrogen = 18.50 g / 14.01 g/mol ≈ 1.32 mol

Oxygen:
moles of oxygen = grams of oxygen / molar mass of oxygen
moles of oxygen = 42.27 g / 16.00 g/mol ≈ 2.64 mol

Step 4: Divide each number of moles by the smallest number of moles calculated in step 3.

The smallest number of moles is 1.32 mol, so divide each number of moles by 1.32:
moles of carbon = 3.08 mol / 1.32 mol ≈ 2.33
moles of hydrogen = 2.20 mol / 1.32 mol ≈ 1.67
moles of nitrogen = 1.32 mol / 1.32 mol = 1
moles of oxygen = 2.64 mol / 1.32 mol ≈ 2

Step 5: Round the resulting ratios to the nearest whole number to get the subscripts in the empirical formula.

The empirical formula is C2.33H1.67N1O2.

Step 6: Multiply all the subscripts by a common factor to obtain whole numbers if necessary. In this case, multiplying by 3 will give whole number coefficients.

3 × C2.33H1.67N1O2 ≈ C7H5N3O6

Therefore, the empirical formula for the given compound is C7H5N3O6, which matches your teacher's answer.

To determine the empirical formula, you need to find the simplified ratio of elements in the compound. Here's how you can calculate the empirical formula step by step:

1. Convert the percentages to grams:
- Assume you have 100 grams of the compound.
- This means you have 37.01 grams of carbon, 2.22 grams of hydrogen, 18.50 grams of nitrogen, and 42.27 grams of oxygen.

2. Convert the grams to moles:
- Divide the grams of each element by their respective molar mass.
- The molar masses are approximately 12.01 g/mol for carbon, 1.01 g/mol for hydrogen, 14.01 g/mol for nitrogen, and 16.00 g/mol for oxygen.
- This gives you approximately:
- Carbon: 37.01 g / 12.01 g/mol ≈ 3.08 moles
- Hydrogen: 2.22 g / 1.01 g/mol ≈ 2.20 moles
- Nitrogen: 18.50 g / 14.01 g/mol ≈ 1.32 moles
- Oxygen: 42.27 g / 16.00 g/mol ≈ 2.64 moles

3. Divide each mole value by the smallest mole value:
- Divide each number of moles by the smallest value obtained, which is approximately 1.32 moles.
- This gives you approximately:
- Carbon: 3.08 moles / 1.32 moles ≈ 2.33
- Hydrogen: 2.20 moles / 1.32 moles ≈ 1.67
- Nitrogen: 1.32 moles / 1.32 moles = 1.00
- Oxygen: 2.64 moles / 1.32 moles ≈ 2.00

4. Round the resulting numbers to the nearest whole number:
- You end up with the numbers: 2.33, 1.67, 1.00, and 2.00
- After rounding, these become: 2, 2, 1, and 2

5. Write the empirical formula using the whole numbers obtained:
- The empirical formula is C2H2N1O2.
- However, you mentioned that your teacher gave you the formula C7H5N3O6. This implies that the compound is a multiple of the empirical formula (C2H2N1O2).

It's important to note that the empirical formula only shows the simplest whole-number ratio of atoms in a compound. If the given compound has a molecular formula different from the empirical formula, it suggests that it is a multiple of the empirical formula.