a compound with the molecular weight of 276 grams was determined to be 39.10% carbon, 8.77% hydrogen, and 52.13% oxygen. what is the empiricial formula and the molecular formula for the compound? i do not know where to start!
i worked it out and i got 276/100=2.76 do i round it up to 3? SO THE MOLECULAR FORMULA IS 3(ch30)? then i divided 39.10/12.01=3.25/3.25 =1 and the same for oxygen becaseu their answers were both the smallest. for hydrogen i divided 8.77/1.008=8.70/3.25=2.67 do you round up to 3? so the EMPIRICAL FORMULA is CH30 because you round up? or do i keep it as CH20? THANKS SO MUCH!!!!
You did great. I would not round 2.67 to 3. I would look for a multiplier where all would be whole numbers; for example, if we multiply by 3 we get C = 3, H = 8.01 (which we round to 8.0), O = 3 so the empirical formula is C3H8O3. That makes the empirical formula mass of 36 + 8 + 48 = 92.
Then 276/92 = 3 so the molecular formula is
(C3H8O3)3 = C9H27O9. Check that to make sure it adds to 276. Generally we round if the number is close to a whole number (2.9 would round to 3 and 2.1 would round to 2.0. Generally, numbers close to 0.25, 0.5 and 0.75 can be multiplied to obtain whole numbers. 4*0.25 = 1. etc.
To determine the empirical formula and molecular formula, you need to follow these steps:
Step 1: Determine the empirical formula.
To find the empirical formula, you need to determine the ratio of the elements present in the compound. Start by assuming you have 100 grams of the compound, which will allow you to convert the percentages into grams.
Given:
Carbon (C) = 39.10%
Hydrogen (H) = 8.77%
Oxygen (O) = 52.13%
Assume you have 100 grams of the compound. Then, you have:
Carbon (C): 39.10 grams
Hydrogen (H): 8.77 grams
Oxygen (O): 52.13 grams
Next, convert each element's mass into the number of moles using their atomic masses.
Atomic mass (C) = 12.01 g/mol
Atomic mass (H) = 1.008 g/mol
Atomic mass (O) = 16.00 g/mol
Moles of C = 39.10 grams / 12.01 g/mol ≈ 3.257 moles
Moles of H = 8.77 grams / 1.008 g/mol ≈ 8.696 moles
Moles of O = 52.13 grams / 16.00 g/mol ≈ 3.258 moles
Now, divide each element's moles by the smallest number of moles to get the ratio.
Dividing by 3.257 (smallest number of moles):
C: 3.257 moles / 3.257 = 1
H: 8.696 moles / 3.257 ≈ 2.667 ≈ 3 (rounding up)
O: 3.258 moles / 3.257 ≈ 1 ≈ 1
Therefore, the empirical formula is CH3O.
Step 2: Determine the molecular formula.
To find the molecular formula, you need to know the molecular weight of the compound.
Given: Molecular weight = 276 grams
Calculate the molar mass of the empirical formula CH3O:
(1 atom of C) + (3 atoms of H) + (1 atom of O) = 12.01 g/mol + (3 × 1.008 g/mol) + 16.00 g/mol = 31.037 g/mol
Next, divide the molecular weight by the molar mass of the empirical formula to find the whole number multiple.
Molecular weight / Empirical formula molar mass:
276 g / 31.037 g/mol ≈ 8.888 ≈ 9
Finally, multiply the empirical formula by the whole number multiple to obtain the molecular formula.
(CH3O) × 9 = C9H27O9
Therefore, the molecular formula is C9H27O9.
Note: It is essential to round the atomic ratios to the nearest whole number when determining the empirical formula. In the case of hydrogen (H), rounding up from 2.667 to 3 is correct.
To determine the empirical formula and the molecular formula for the compound, you need to follow a step-by-step process. Let's break it down together:
1. Start by assuming you have 100 grams of the compound. This assumption simplifies the calculations, as percentages can be directly converted into grams.
2. Convert the percentages of carbon, hydrogen, and oxygen into grams. Given that the compound is 39.10% carbon, 8.77% hydrogen, and 52.13% oxygen, you would have:
- Carbon: 100g * 0.3910 = 39.10g
- Hydrogen: 100g * 0.0877 = 8.77g
- Oxygen: 100g * 0.5213 = 52.13g
Now, we have the gram amounts for each element.
3. Convert the grams of each element into moles by dividing by their respective molar masses:
- Carbon: 39.10g / 12.01g/mol = 3.257 moles
- Hydrogen: 8.77g / 1.008g/mol = 8.7 moles
- Oxygen: 52.13g / 16.00g/mol = 3.257 moles
4. Divide each number of moles by the smallest value obtained. In this case, the smallest value is 3.257 moles. Divide each number of moles by 3.257:
- Carbon: 3.257 moles / 3.257 moles = 1 mole
- Hydrogen: 8.7 moles / 3.257 moles = 2.67 moles (rounded to 2)
- Oxygen: 3.257 moles / 3.257 moles = 1 mole
5. Based on the whole number ratio obtained from dividing by the smallest value, determine the empirical formula. The empirical formula represents the simplest, most reduced ratio of atoms in the compound. In this case, it would be CH2O.
Now, for the molecular formula:
6. Calculate the empirical formula's molar mass by adding up the molar masses of its components:
- Carbon: 1 * 12.01g/mol = 12.01g/mol
- Hydrogen: 2 * 1.008g/mol = 2.016g/mol
- Oxygen: 1 * 16.00g/mol = 16.00g/mol
Therefore, the empirical formula's molar mass is 30.03g/mol.
7. Divide the given molecular weight (276g) by the empirical formula's molar mass (30.03g/mol) to determine the whole number ratio between the molecular formula and the empirical formula:
- 276g / 30.03g/mol = 9.19
8. Round the whole number ratio obtained to the nearest whole number. In this case, round 9.19 to 9.
9. Multiply the empirical formula by the rounded whole number to obtain the molecular formula:
- Empirical formula: CH2O
- Multiply by 9: 9 * CH2O = C9H18O9
Therefore, the empirical formula for the compound is CH2O, and the molecular formula is C9H18O9.
Note: In this particular case, rounding to the nearest whole number resulted in a molecular formula that is a multiple of the empirical formula. However, it's important to note that this may not always be the case. Sometimes the empirical and molecular formulas can be the same, while other times they may differ.