when hydrocarbons are burned in a limited amount of air CO as well as CO2 form. When 0.450 grams of a particular hydrocarbon was burned in air, 0.467 grams of CO, 0.733 grams of CO2, and 0.450 grams of H20 were formed.

a. What is the empirical formula?
I got C2H3! check?

b. How many grams of O2 were used in the reaction?

c. How many grams were required for complete combustion?

Thanks!

a. C2H3 is incorrect. To determine the empirical formula, we need to find the moles of each element.

Moles of C in CO: 0.467 g / 28.01 g/mol = 0.0166 mol
Moles of C in CO2: 0.733 g / 44.01 g/mol = 0.0166 mol (Notice that this is the same as CO)
Moles of C in hydrocarbon: 0.450 g / molar mass of C

Similarly, for H2:
Moles of H in H2O: 0.450 g / 18.02 g/mol = 0.0249 mol

From the ratios, it seems like the empirical formula would be:

C: 0.0166 mol / 0.0166 mol = 1
H: 0.0249 mol / 0.0166 mol = 1.5

To convert the H to a whole number, we multiply both elements by 2:
C: 2
H: 3

Therefore, the empirical formula is C2H3.

b. To find the grams of oxygen used, we need to determine the difference in the mass of the products and reactants. In this case, the only reactant is the hydrocarbon:

Mass of O2 used = Mass of hydrocarbon - Mass of CO2 produced = 0.450 g - 0.733 g = -0.283 g

We obtained a negative value, indicating an error in the calculations. Please double-check your given data or re-calculate.

c. To find the grams of oxygen required for complete combustion, we must assume that the oxygen reacts with all the carbon in the hydrocarbon to form CO2:

Molar mass of C: 12.01 g/mol
Molar mass of CO2: 44.01 g/mol

Moles of C in the hydrocarbon: 0.450 g / 12.01 g/mol
Moles of O2 required: Moles of C x (1 mol O2 / 1 mol C) = Moles of C x (32.00 g/mol O2 / 12.01 g/mol C)

Mass of O2 required: Moles of O2 required x Molar mass of O2

Please re-check your given data or provide additional information to proceed with calculation c.

a. To determine the empirical formula, we need to find the smallest whole number ratio of the elements present in the compound.

First, calculate the moles of each element present in the products:

Moles of CO = mass of CO / molar mass of CO = 0.467 g / 28.01 g/mol = 0.0167 mol
Moles of CO2 = mass of CO2 / molar mass of CO2 = 0.733 g / 44.01 g/mol = 0.0166 mol
Moles of H2O = mass of H2O / molar mass of H2O = 0.450 g / 18.02 g/mol = 0.0249 mol

Next, we need to determine the mole ratio by dividing each of the moles obtained by the smallest moles value:

Ratio of CO:CO2:H2O = 0.0167 mol / 0.0166 mol / 0.0249 mol = 1 : 0.993 : 1.498

Since we are looking for whole numbers, we can round the ratio to the nearest whole number:

Ratio of CO:CO2:H2O = 1 : 1 : 1.5

The empirical formula of the hydrocarbon is C2H3O.

b. To calculate the number of grams of O2 used in the reaction, we need to subtract the masses of CO and CO2 formed from the total mass of the products:

Mass of O2 = Total mass of products - (Mass of CO + Mass of CO2) = (0.467 g + 0.733 g + 0.450 g) - (0.467 g + 0.733 g) = 0.450 g

Therefore, 0.450 grams of O2 were used in the reaction.

c. To determine the grams of O2 required for complete combustion, we need to assume complete conversion of the hydrocarbon to CO2 and H2O. In this case, the mole ratio between CO2 and O2 is 2:1, so the number of moles of O2 required is twice the number of moles of CO2:

Moles of CO2 = 0.0166 mol
Moles of O2 required = 2 * Moles of CO2 = 2 * 0.0166 mol = 0.0332 mol

Now, calculate the grams of O2 required:

Mass of O2 required = Moles of O2 required * molar mass of O2 = 0.0332 mol * 32.00 g/mol = 1.06 g

Therefore, 1.06 grams of O2 is required for complete combustion.

a. To find the empirical formula, we need to determine the ratio of the elements in the compound.

First, we need to find the moles of each element in the given compounds:

- Moles of CO: 0.467 grams of CO / molar mass of CO (28.010 g/mol) = 0.01665 moles of CO
- Moles of CO2: 0.733 grams of CO2 / molar mass of CO2 (44.009 g/mol) = 0.01665 moles of CO2
- Moles of H2O: 0.450 grams of H2O / molar mass of H2O (18.015 g/mol) = 0.02498 moles of H2O

From the moles of each element, we can determine the molar ratios. We divide each mole value by the smallest mole value:

- CO: 0.01665 moles / 0.01665 moles = 1
- CO2: 0.01665 moles / 0.01665 moles = 1
- H2O: 0.02498 moles / 0.01665 moles = 1.5

Now, we need to convert these ratios to whole numbers by multiplying each ratio by the smallest possible factor that will give whole numbers. In this case, the smallest factor is 2:

- CO: 1 * 2 = 2
- CO2: 1 * 2 = 2
- H2O: 1.5 * 2 = 3

Therefore, the empirical formula is C2H3.

b. To find the grams of O2 used in the reaction, we use the fact that the hydrocarbon is burned completely. This means that all the carbon in the hydrocarbon combines with oxygen to form CO2.

The molar mass of CO2 is 44.009 g/mol, and we have 0.733 grams of CO2. So, we can calculate the moles of CO2:

- Moles of CO2: 0.733 grams of CO2 / molar mass of CO2 (44.009 g/mol) = 0.01665 moles of CO2

According to the balanced chemical equation for the combustion of hydrocarbons, one mole of CO2 requires one mole of O2. Therefore, the moles of O2 used in the reaction are also 0.01665 moles.

To find the grams of O2, we multiply the moles of O2 by the molar mass of O2:

- Grams of O2: 0.01665 moles of O2 * molar mass of O2 (32 g/mol) = 0.5328 grams of O2

Therefore, approximately 0.5328 grams of O2 were used in the reaction.

c. To determine the grams of O2 required for complete combustion, we use the mole ratio between carbon and O2 in the balanced chemical equation for combustion.

From the empirical formula C2H3, we can determine that there are 2 moles of carbon in the hydrocarbon. According to the balanced equation, each mole of carbon requires one mole of O2 for complete combustion.

Therefore, we need 2 moles of O2 for every 2 moles of carbon. Calculating the molar mass of O2 gives us 32 g/mol.

- Grams of O2 required: 2 moles of O2 * molar mass of O2 (32 g/mol) = 64 grams of O2

Therefore, 64 grams of O2 are required for complete combustion.