# Molecular formula of combustion?

• Chemistry
• raoulduke1
In summary, complete combustion of 1.1g of a compound of C, H, and O produces 2.2g of CO2 and 0.9g of H2O, with a relative molar mass close to 130. To find the molecular formula and true molar mass, divide the mass of each substance by its molar mass. In this case, the number of moles of CO2 is 0.05. There may be a discrepancy in the mass due to the presence of oxygen in combustion.
raoulduke1
complete combustion of 1.1g of a compound of C,H and O gives 2.2g of CO2 and 0.9g of H2O.
The compound has a relative molar mass close to 130.
Find the molecular formula and true molar mass?

dont even no where to begin

so you got the masses of the products right? 2.2g for CO2 and 0.9g for H2O.

you knowthe molar masses of those elements by looking at the periodic table

for example, the moalr mass of CO2 is (approximately because of rounding)= 12+16+16=44g/moldo the same thing to get the molar mass of H2O (hydrogen is 1) to get the H2O moles

what has me confused though is that it doesn't appear to obbey the conservation of mass law, how can you have 1.1g of something and get 2.2g+0.9g of something else?maybe I am just reading wrong...

then divide the mass of the substances by the molar massses, to get number of moles, so 2.2/44=0.05 moles of CO2

stonecoldgen said:
what has me confused though is that it doesn't appear to obbey the conservation of mass law, how can you have 1.1g of something and get 2.2g+0.9g of something else?

maybe I am just reading wrong...

Combustion means there is a source of oxygen.

Please pay attention to what and how you write, don't ignore capital letters and watch for typos.

I would first start by calculating the molar masses of CO2 and H2O using the atomic masses of carbon, hydrogen, and oxygen. From the given information, we know that 1.1g of the compound produced 2.2g of CO2 and 0.9g of H2O. Using the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol), we can calculate the moles of each compound produced.

2.2 g CO2 x (1 mol CO2 / 44.01 g CO2) = 0.05 mol CO2
0.9 g H2O x (1 mol H2O / 18.02 g H2O) = 0.05 mol H2O

Since the ratio of moles of CO2 and H2O is 1:1, we can infer that the number of moles of carbon and hydrogen in the compound must also be in a 1:1 ratio. This means that the molecular formula of the compound is likely C1H1, which simplifies to CH.

To find the true molar mass, we can use the given information that the relative molar mass is close to 130. This means that the actual molar mass is likely slightly higher than 130 g/mol. Since the molecular formula is CH, the molar mass would be the sum of the atomic masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol), which is 13.02 g/mol. This is significantly lower than 130 g/mol, so it is possible that the compound has a higher molecular formula, such as C6H6, which would have a molar mass of 78.11 g/mol.

In conclusion, based on the given information, the molecular formula of the compound is likely CH, and the true molar mass is possibly higher than 130 g/mol, potentially closer to 78.11 g/mol. Further analysis and experimentation may be needed to determine the exact molecular formula and molar mass of the compound.

## 1. What is the molecular formula of combustion?

The molecular formula of combustion is CxHyOz, where x, y, and z represent the number of carbon, hydrogen, and oxygen atoms in a molecule, respectively.

## 2. How is the molecular formula of combustion determined?

The molecular formula of combustion is determined by balancing the chemical equation for the combustion reaction. This involves counting the number of each type of atom on both the reactant and product sides of the equation and adjusting the coefficients to ensure that the same number of each type of atom is present on both sides.

## 3. How does the molecular formula of combustion relate to energy release?

The molecular formula of combustion is directly related to the amount of energy released during the combustion reaction. This is because the combustion reaction involves the breaking of bonds between atoms in the fuel molecule, followed by the formation of new bonds between atoms in the combustion products. The energy released is determined by the types of atoms and the number and strength of bonds involved.

## 4. Can the molecular formula of combustion vary?

Yes, the molecular formula of combustion can vary depending on the type of fuel being burned. Different fuels have different compositions, leading to different molecular formulas for their combustion reactions.

## 5. How is the molecular formula of combustion used in practical applications?

The molecular formula of combustion is used in practical applications such as determining the efficiency of a fuel source, predicting the amount of energy released during a combustion reaction, and developing new and more efficient fuels for various applications.

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