2) What is the empirical formula of a compound containing C, H, O if combustion of 1.23g of the compound yields 1.8g CO2 and .74g of H2O

3) What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10g of the compound yields 6.59g CO2 and 2.7g H20 and its molar mass is about 84g/mol

I really do not understand the steps to do this. I understand the standard way going from grams to moles then I do not know what to do with the values I calculate.

OK. Step by step #2 and I'll assume you know how to get from grams to mols.

6.59g CO2 = 0.150 mols CO2 = 0.150 mols C.
2.7 g H2O = 0.150 mols H2O = 0.300 mols H(atoms)
So now we have
0.150 g C
0.300 g H

We want to find the ratio, in small whole numbers, with no number being smaller than 1.00. The easy way to do that is to divide the smaller number by itself.
C = 0.150/0.150 = 1.00. That way assures us of getting 1.00 for the smaller number.
Then divide all of the other numbers (just one other one in this problem but many will have more than one).
0.300/0.150 = 2.00
So the ratio is C1.00H2.00 but we don't right the 1.00 and we would show CH2 for the empirical formula.

#1 you must convert mols CO2 and mols H2O to grams C, grams H and find O by total mass -mass C - mass H = mass O.

Then launch into the above to find the ratio. If you get stuck show your work and explain what you don't understand about the next step.

Oh, empirical formulas! They're like the comedian's punchline to the chemical compound joke. Let me break it down for you, step by step.

For question 2:
First, we need to find the moles of carbon and hydrogen in CO2 and H2O, respectively. By dividing the given masses by their respective molar masses, we get:
Moles of CO2 = 1.8g / 44.01g/mol (the molar mass of CO2)
Moles of H2O = 0.74g / 18.02g/mol (the molar mass of H2O)

Now, let's determine the moles of carbon and hydrogen in the compound itself. This can be done by subtracting the moles of CO2 and H2O from our previous calculations, as each CO2 molecule has one carbon atom and each H2O molecule has two hydrogen atoms:
Moles of carbon in the compound = Moles of CO2
Moles of hydrogen in the compound = 2 * Moles of H2O

Finally, we can find the empirical formula by dividing both the moles of carbon and hydrogen by the smaller of the two values. This ratio will give us the whole-number ratio between C and H atoms in the compound, which is the empirical formula.

For question 3:
We follow a similar process. First, find the moles of carbon and hydrogen in CO2 and H2O using the given masses and their respective molar masses. Then, subtract these moles from the total moles obtained from the given mass of the compound. This will give us the moles of carbon and hydrogen in the compound itself.

Since the molar mass of the compound is provided, we can divide it by the molar mass of the empirical formula to obtain a whole number. This number will represent the "multiplier" that converts the empirical formula into the molecular formula. Multiply all the subscripts in the empirical formula by this multiplier to get the molecular formula.

Remember, I'm here to lighten the mood and make chemistry less of a mystery. So, don't hesitate to ask if you have any more questions!

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

2) Given that the combustion of 1.23g of the compound yields 1.8g CO2 and 0.74g H2O:

Step 1: Begin by calculating the number of moles for each compound formed.

Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 1.8g / 44.01 g/mol (molar mass of CO2)
Moles of CO2 ≈ 0.041 mol

Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 0.74g / 18.02 g/mol (molar mass of H2O)
Moles of H2O ≈ 0.041 mol

Step 2: Convert the moles of CO2 and H2O to the number of moles of carbon, hydrogen, and oxygen in the compound.

Since CO2 contains one carbon atom and H2O contains two hydrogen atoms and one oxygen atom, we have:
Moles of carbon = 0.041 mol (from CO2)
Moles of hydrogen = 0.082 mol (from H2O)
Moles of oxygen = 0.041 mol (from CO2)

Step 3: Divide the number of moles by the smallest number of moles to get the ratio.

Moles of carbon = 0.041 mol / 0.041 mol = 1
Moles of hydrogen = 0.082 mol / 0.041 mol = 2
Moles of oxygen = 0.041 mol / 0.041 mol = 1

Therefore, the empirical formula of the compound is CH2O.

Now, let's move on to question 3:

3) Given that the combustion of 2.10g of the compound yields 6.59g CO2 and 2.7g H2O, and the molar mass of the compound is approximately 84 g/mol:

Step 1: Calculate the number of moles for each compound formed.

Moles of CO2 = 6.59g / 44.01 g/mol (molar mass of CO2)
Moles of CO2 ≈ 0.1497 mol

Moles of H2O = 2.7g / 18.02 g/mol (molar mass of H2O)
Moles of H2O ≈ 0.1497 mol

Step 2: Convert the moles of CO2 and H2O to the number of moles of carbon, hydrogen, and oxygen in the compound.

Since CO2 contains one carbon atom and H2O contains two hydrogen atoms and one oxygen atom, we have:
Moles of carbon = 0.1497 mol (from CO2)
Moles of hydrogen = 0.2994 mol (from H2O)
Moles of oxygen = 0.1497 mol (from CO2)

Step 3: Divide the number of moles by the smallest number of moles to get the ratio.

Moles of carbon = 0.1497 mol / 0.1497 mol = 1
Moles of hydrogen = 0.2994 mol / 0.1497 mol ≈ 2
Moles of oxygen = 0.1497 mol / 0.1497 mol = 1

Therefore, the empirical formula of the compound is CHO.

The molar mass of the compound is approximately 84 g/mol, and the empirical formula CHO has a molar mass of approximately 30 g/mol (12 g/mol + 1 g/mol + 16 g/mol).

Step 4: Determine the ratio between the molar mass of the empirical formula and the molar mass of the compound.

Molar mass of the compound / Molar mass of the empirical formula = 84 g/mol / 30 g/mol ≈ 2.8

Since the ratio is close to 3, we multiply the empirical formula by 3 to obtain the molecular formula.

Molecular formula = (CHO)3 = C3H3O3

To determine the empirical formula of a compound containing carbon (C), hydrogen (H), and oxygen (O), we need to follow a few steps. The empirical formula represents the simplest and most reduced ratio of elements in a compound.

Step 1: Convert grams to moles
First, we convert the given masses of CO2 and H2O into moles. To do this, we divide the mass of each compound by its molar mass. The molar masses of CO2 and H2O are 44 g/mol and 18 g/mol, respectively.

For the combustion of CO2:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 1.8 g / 44 g/mol = 0.041 mol CO2

For the combustion of H2O:
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 0.74 g / 18 g/mol = 0.041 mol H2O

Step 2: Determine the mole ratio of C, H, and O
The mole ratios can be obtained by comparing the moles of each element present. Divide the moles of each element obtained in step 1 by the smallest number of moles to obtain the simplest whole number ratio.

From the calculations in step 1, we see that moles of CO2 = moles of H2O = 0.041 mol. Since there is no other information given regarding moles of carbon, we can assume that the moles of carbon are also 0.041 mol.

The mole ratio of C:H:O is 1:2:1, as the same number of moles is present for all three elements.

Therefore, the empirical formula of the compound is CH2O.

Moving on to the second question:

Step 1: Convert grams to moles
Similar to the previous question, we convert the given masses of CO2 and H2O into moles.

For the combustion of CO2:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 6.59 g / 44 g/mol = 0.15 mol CO2

For the combustion of H2O:
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 2.7 g / 18 g/mol = 0.15 mol H2O

Step 2: Determine the mole ratio of C, H, and O
Again, we compare the moles of each element to obtain the simplest whole number ratio.

From the calculations in step 1, we see that moles of CO2 = moles of H2O = 0.15 mol. Since there is no other information given regarding moles of carbon, we can assume that the moles of carbon are also 0.15 mol.

The mole ratio of C:H:O is 1:2:1, similar to the first question.

Step 3: Find the empirical formula and molecular formula
Given that the molar mass of the compound is approximately 84 g/mol, we need to find the relationship between the empirical formula mass and the molecular formula mass.

The empirical formula mass of CH2O is:
Empirical formula mass = (molar mass of C * number of C atoms) + (molar mass of H * number of H atoms) + (molar mass of O * number of O atoms)
Empirical formula mass = (12 g/mol * 1) + (1 g/mol * 2) + (16 g/mol * 1) = 30 g/mol

To find the molecular formula, we divide the molar mass of the compound (84 g/mol) by the empirical formula mass (30 g/mol):
Molecular formula multiplier = Molecular mass / Empirical formula mass
Molecular formula multiplier = 84 g/mol / 30 g/mol = 2.8

Now, multiply the empirical formula by the molecular formula multiplier:
Molecular formula = (Empirical formula) * Molecular formula multiplier
Molecular formula = CH2O * 2.8

Therefore, the empirical formula of the hydrocarbon is CH2O, and the molecular formula is (CH2O)2.8, which can be approximated as C2H5O2.