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Cumene is the common name the IUPAC standard name is Isopropylbenzene
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Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water?
The molar mass of H2O = 2 + 16 = 18 g/mol. The ratio of H in H2O is therefore 2/18 = 1/9. Thus, amount of H in 42.8 mg H2O = 42.8 * 1/9 = 4.76 mg. This must then be the amount of H in the 47.6 mg cumene. Subtracting, 47.6 - 4.76 = 42.84 mg, is therefore the amount of C in the cumene. Summarising: 47.6 mg cumene = 42.84 mg C + 4.76 mg H Converting mass of C and H to millimoles, we have : C = 42.84 /12 = 3.57 mmol and H = 4.76/1 = 4.76 mmol. The molar ratio C:H is therefore 3.57:4.76. Dividing each by the smallest (that is, 3.57) gives : C:H = 1:1.33 Now, multiplying each by 3 brings us to the integers : C:H = 3:4 So the empirical formula for cumene is C3H4. This has a molar mass of 3*12 + 4*1 = 40 g/mol. Given that the molar mass is between 115 g/mol and 125 g/mol, if we multiply 40 by 3, this gives 120 g/mol, smack bang in the middle. So, molecular formula of cumene is C9H12.
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The chemical formula of cumene is C9H12.
What is the Reaction of cumene and acetyl chloride in the presence of aluminum chloride?
The name of this reaction is Friedel-Crafts acylation. It should have a major product adding a ketone at the para position with regard to the isopropyl of the cumene.
The empirical formula for cumene a compound containing only carbon and hydrogen is C3H4. Its Molecular mass is 120 g. What is the molecular formula for cumene?
This molecular formula is C9H12.
How much does one gallon of cumene weigh?
7.213 lbs
Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water?
The molar mass of H2O = 2 + 16 = 18 g/mol. The ratio of H in H2O is therefore 2/18 = 1/9. Thus, amount of H in 42.8 mg H2O = 42.8 * 1/9 = 4.76 mg. This must then be the amount of H in the 47.6 mg cumene. Subtracting, 47.6 - 4.76 = 42.84 mg, is therefore the amount of C in the cumene. Summarising: 47.6 mg cumene = 42.84 mg C + 4.76 mg H Converting mass of C and H to millimoles, we have : C = 42.84 /12 = 3.57 mmol and H = 4.76/1 = 4.76 mmol. The molar ratio C:H is therefore 3.57:4.76. Dividing each by the smallest (that is, 3.57) gives : C:H = 1:1.33 Now, multiplying each by 3 brings us to the integers : C:H = 3:4 So the empirical formula for cumene is C3H4. This has a molar mass of 3*12 + 4*1 = 40 g/mol. Given that the molar mass is between 115 g/mol and 125 g/mol, if we multiply 40 by 3, this gives 120 g/mol, smack bang in the middle. So, molecular formula of cumene is C9H12.
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