What is the empirical formula of a compound that is 92.3 carbon and 7.7 hydrogen?
Table of Contents
What is the empirical formula of a compound that is 92.3 carbon and 7.7 hydrogen?
A hydrocarbon contains 92.3% by mass of carbon and 7.7% by mass of hydrogen. The empirical formula of the compound will be oose answer: CH CH5 CH CH3.
What is the molecular formula of a compound with 92.3% carbon and 7.7% hydrogen and a molecular mass of 78?
The molecule of the compound is 39 times heavier than hydrogen molecule. The molecular formula of the compound is 11 C H B C D 3) C H 4 C H.
What is the molecular formula for a compound that is 92.24 carbon and 7.76% hydrogen and has a molar mass of 78.12 g *?
Both Benzene (C6H6, molar mass = 78.12g/mol) and acetylene (C2H2, molar mass = 26.04g/mol) have the same percent composition (92.24 mass% carbon and 7.76% hydrogen) and the empirical formula, CH.
What is the empirical formula for carbon and hydrogen?
CH2
Its empirical formula is CH2. One molecule of ethylene (molecular formula C2H4) contains two atoms of carbon and four atoms of hydrogen. Its empirical formula is CH2….
| Empirical formula | CH (92.2% C; 7.8% H) |
|---|---|
| Compound | acetylene |
| Molecular formula | C2H2 |
| Boiling point, °C | -84 |
How do you find the molecular formula from the empirical formula?
Divide the molar mass of the compound by the empirical formula molar mass. The result should be a whole number or very close to a whole number. Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.
What is the empirical formula of a compound that contains 7.69 of H and 92.3 carbon?
a compound of molecular mass 78 gm contains 92.3% carbon and 7.7% hydrogen. so mass of carbon in that compound = 78×92.3/100=72gm. and mass of hydrogen in that compound = 78×7.7/100= 6 gm. so the molecular formula is C6H6.
What is the simplest formula of the compound which has C 92.31 and H 7.69 if the molecular mass is 78 C calculate its molecular formula?
Empirical formula of benzene `{:(“Element”,”Percentage”,”Atomic mass”,”Gram atoms (Moles)”,”Atomic ratio (Molar ratio)”,”Simplest whole no. ratio”),(“C”,92.31,12,(93.31)/(12)=7.69,(7.69)/(7.69)=1,1),(“H”,7.69,1,(7.69)/(1)=7.69,(7.69)/(7.69)=1,1):}` Empirical formula of benzene = CH Step II.
What is the empirical formula of a compound that has 40% carbon 6.7% hydrogen and 53.3% oxygen?
1 Answer. Ernest Z. The empirical formula is CH2O .
How do you convert an empirical formula to a molecular formula?
What are the steps to finding the molecular formula?
The steps involved in determining a Molecular Formula are:
- Find the mass of the empirical formula.
- Divide the molecular mass by the mass of the empirical formula.
- Multiply each subscript in the empirical formula by the answer in step 2.
What is the molecular formula of a compound with the empirical formula CH and a formula mass of 78.114 amu?
What is the molecular formula of a compound with the empirical formula CH and a formula mass of 78.114 amu? molecular formula: CzHz What is the molecular formula of a compound with the empirical formula CH,C 0 and a formula mass of 90.078 amu?
What is the empirical formula of a compound that has 40.1% C 6.6% H and 53.3% O?
The empirical formula is CH2O .
What is the empirical formula of a compound that is 40.00% C 6.72% H and 53.29% O by weight?
The empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆. Assume we have 100 g of the compound. Then we have 40.00 g of C, 6.72 g of H, and 53.28 g of O. The empirical formula is CH₂O.
