Calculate the empirical formula of Tylenol: C, 63.56%; H, 6.00%; N, 9.27%; O, 21.17%
Follow the sequence:
% => gms => moles => reduce moles (divide by smallest mole value) => Emp. Ratio => C:H:N:O
C: 63.56% => 63.56g/12g/mol = 5.297mol
H: 6.00% => 6.00g/ 1g/mol = 6.000mol
N: 9.27% => 9.27g/14g/mol = 0.662mol
O: 21.17% => 21.17g/16g/mol = 1.323mol
N:H:N:O Emp Ratio =>
5.297 6.000 0.662 1.323
_____ : _____ : _____ : _____
0.662 0.662 0.662 0.662
=> 8:9:1:2 Empirical Ratio
=> C8H9NO2
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Sure, let's calculate the empirical formula of Tylenol, with a dash of humor added for good measure!
So, to find the empirical formula, we need to determine the ratio of atoms in the compound. Let's start by assuming we have 100 grams of Tylenol.
Now, we have:
- Carbon (C) = 63.56 grams
- Hydrogen (H) = 6.00 grams
- Nitrogen (N) = 9.27 grams
- Oxygen (O) = 21.17 grams
Now, let's take a look at these numbers.
As you can see, there's quite a party going on in Tylenol! Carbon is hitting the dance floor at 63.56 grams, Hydrogen decided to do some low-key dancing at 6.00 grams, Nitrogen is grooving to its own beat at 9.27 grams, and Oxygen is trying to keep up with its 21.17 grams.
Now, let's convert these numbers into moles. Remember, moles are like the disco balls of chemistry!
- Carbon (C): 63.56 grams / 12.01 g/mol = 5.289 moles
- Hydrogen (H): 6.00 grams / 1.01 g/mol = 5.941 moles
- Nitrogen (N): 9.27 grams / 14.01 g/mol = 0.662 moles
- Oxygen (O): 21.17 grams / 16.00 g/mol = 1.323 moles
Now, let's find the mole-to-mole ratio by dividing the number of moles by the smallest number of moles, which is Nitrogen in this case.
So, Carbon (C) becomes 5.289 / 0.662 ≈ 8
Hydrogen (H) becomes 5.941 / 0.662 ≈ 9
Nitrogen (N) stays the same
Oxygen (O) becomes 1.323 / 0.662 ≈ 2
Now, let's write the empirical formula using these ratios:
C8H9N2O2
And there you have it! The empirical formula of Tylenol is C8H9N2O2. It seems like Tylenol sure knows how to have a chemical party!
To determine the empirical formula of Tylenol, we need to know the percentage composition of each element present in the compound. In this case, we have the following percentages:
C (carbon) - 63.56%
H (hydrogen) - 6.00%
N (nitrogen) - 9.27%
O (oxygen) - 21.17%
To calculate the empirical formula, we need to convert these percentages into moles. The mole ratio between the elements in a compound will give us the simplest whole-number ratio of the elements. Let's go step by step:
Step 1: Convert the percentages to grams.
Assume we have 100 grams of the compound, which would mean:
- Carbon: 63.56 grams
- Hydrogen: 6.00 grams
- Nitrogen: 9.27 grams
- Oxygen: 21.17 grams
Step 2: Convert grams to moles.
To convert grams to moles, we divide the mass in grams by the molar mass of each element.
The molar mass for each element:
- C (carbon) = 12.01 g/mol
- H (hydrogen) = 1.01 g/mol
- N (nitrogen) = 14.01 g/mol
- O (oxygen) = 16.00 g/mol
Now, divide the grams by the molar masses to get the moles for each element:
- Carbon moles (C) = 63.56 g / 12.01 g/mol
- Hydrogen moles (H) = 6.00 g / 1.01 g/mol
- Nitrogen moles (N) = 9.27 g / 14.01 g/mol
- Oxygen moles (O) = 21.17 g / 16.00 g/mol
After calculating these values, we find:
- Carbon moles (C) ≈ 5.29 mol
- Hydrogen moles (H) ≈ 5.94 mol
- Nitrogen moles (N) ≈ 0.66 mol
- Oxygen moles (O) ≈ 1.32 mol
Step 3: Find the simplest whole-number ratio by dividing all the mole values by the smallest mole value.
In this case, the smallest mole value is that of nitrogen (N), which is approximately 0.66 mol. Divide all the moles by 0.66 to obtain the simplest whole-number ratio:
- Carbon: 5.29 mol / 0.66 mol ≈ 8
- Hydrogen: 5.94 mol / 0.66 mol ≈ 9
- Nitrogen: 0.66 mol / 0.66 mol = 1
- Oxygen: 1.32 mol / 0.66 mol ≈ 2
The simplest whole-number ratio of carbon, hydrogen, nitrogen, and oxygen in Tylenol (acetaminophen) is approximately C8H9NO2. Therefore, the empirical formula of Tylenol is C8H9NO2.