Which of the following compounds has a chemical formula CH2O?

Q: How do you find the molecular formula of CH2O?

Lets find the Molar mass of Carbon=12..

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The empirical formula sometimes called the simplest or reduced formula and so were given the empirical formula c, h, 2, o so for the first option. Here we have c h: 3, o c h, 3. So this is equivalent to c 2 h, 6 o. And if i divide this by 2, i would get c h: 3, o half, so that's not going to work. Next is c h, 3 c, o o h. This is equivalent to c 2 h. 4. O 2, divided in half would give me c h. 2. O so that works, the next 1 is c h, 3 c h, 2, o h. This is equivalent to c 2 h, 6, o c 2 h 6, o that would reduce divided by 2 to c h, 3, o a half. So that doesn't work and the last 1 is c: 6 h, 5. Oh, that's, equivalent to c 6 h. 6. O divide this by 2 would give me c: 3 h, 3, o half that doesn't work. So my answer here has got to be the second option, which would give me an empirical formula of c h: 2, o.

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Molecular Formulas
Which compounds have empirical formula CH2O?
Is C6H12O6 same as CH2O?
Is CH2O acetic acid?
How do you find the molecular formula of CH2O?

Video Transcript

An organic compound has a molar mass of 60 grams per mole and the empirical formula CH2O. If the molar mass of CH2O is 30 grams per mole, what is the molecular formula of the organic compound?

A molecular formula is a chemical formula that expresses the exact number of atoms of each element in a molecule. For example, the organic compound but-1-ene has a molecular formula of C4H8, meaning that a molecule of but-1-ene contains four atoms of carbon and eight atoms of hydrogen. An empirical formula is a chemical formula that expresses the simplest whole-number ratio of atoms of each element in a molecule or ionic compound. The ratio of carbon atoms to hydrogen atoms in a molecule but-1-ene is four to eight. Since an empirical formula expresses the simplest whole-number ratio of atoms of each element, we can divide each of the subscripts by their greatest common factor to simplify the ratio and determine that the empirical formula of but-1-ene is CH2.

So to go from the molecular formula of but-1-ene to the empirical formula, we divide each subscript in the molecular formula by four. And if we wanted to go from the empirical formula of but-1-ene to the molecular formula, we would multiply each subscript in the empirical formula by four. In this question, we know we have an organic compound that has an empirical formula of CH2O. We want to determine the molecular formula from the example. We know that we can get the molecular formula from the empirical formula by multiplying each of the subscripts in the empirical formula by a certain value. But what value should we use?

We can see in the question that we’re given some additional information. We are told the molar mass of the organic compound is 60 grams per mole and the molar mass of CH2O, the empirical formula, is 30 grams per mole. So let’s take another look at but-1-ene to see how the molar masses of the molecular formula and empirical formula are related. The molar mass of the empirical formula of but-1-ene is four times smaller than the molar mass of its molecular formula. In other words, the molar mass of the molecular formula is four times larger than the molar mass of the empirical formula. So if we know the relationship between the molar mass of the empirical formula and the molar mass of the molecular formula, we’ll also know what value to multiply the empirical formula subscripts by to get the molecular formula.

If we divide 60 grams per mole by 30 grams per mole, we find that the molar mass of the organic compound is two times larger than the molar mass of its empirical formula. So to find the molecular formula, we’ll need to multiply each of those subscripts in the empirical formula including the unwritten ones by two. This gives us a molecular formula of C2H4O2.

So an organic compound that has a molar mass of 60 grams per mole and an empirical formula of CH2O has a molecular formula of C2H4O2.

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Learning Objectives

Understand the difference between empirical formulas and molecular formulas.
Determine molecular formula from percent composition and molar mass of a compound.

Below, we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people can distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way to tell glucose and sucrose apart is to determine the molar masses—this approach allows you to easily tell which compound is which.

Figure \(\PageIndex{1}\): (A) the molecular structure of glucose and (B) the molecular structure of sucrose.

Molecular Formulas

Molecular formulas give the kind and number of atoms of each element present in the molecular compound. In many cases, the molecular formula is the same as the empirical formula. The chemical formula will always be some integer multiple (\(n\)) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).

\[\text{ Molecular Formula} = n (\text{Empirical formula}) \nonumber \]

therefore

\[ n = \dfrac{\text{Molecular Formula}}{\text{Empirical Formula}} \nonumber \]

The integer multiple, n, can also be obtained by dividing the molar mass, \(MM\), of the compound by the empirical formula mass, \(EFM\) (the molar mass represented by the empirical formula).

\[ n = \dfrac{MM ( molar mass)}{EFM (empirical formula molar mass)} \nonumber \]

Table \(\PageIndex{1}\) shows the comparison between the empirical and molecular formula of methane, acetic acid, and glucose, and the different values of n. The molecular formula of methane is \(\ce{CH_4}\) and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is \(\ce{C_2H_4O_2}\). Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is \(\ce{C_6H_{12}O_6}\). The structures of both molecules are shown in Figure \(\PageIndex{2}\). They are very different compounds, yet both have the same empirical formula of \(\ce{CH_2O}\).

Table \(\PageIndex{1}\): Molecular Formula and Empirical Formula of Various Compounds.

Name of Compound	Molecular Formula	Empirical Formula	n
Methane	\(\ce{CH_4}\)	\(\ce{CH_4}\)	1
Acetic acid	\(\ce{C_2H_4O_2}\)	\(\ce{CH_2O}\)	2
Glucose	\(\ce{C_6H_{12}O_6}\)	\(\ce{CH_2O}\)	6

Figure \(\PageIndex{2}\): Acetic acid (left) has a molecular formula of \(\ce{C_2H_4O_2}\), while glucose (right) has a molecular formula of \(\ce{C_6H_{12}O_6}\). Both have the empirical formula \(\ce{CH_2O}\).

Empirical formulas can be determined from the percent composition of a compound as discussed in section 6.8. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:

Calculate the empirical formula molar mass (EFM).
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.

Example \(\PageIndex{1}\)

The empirical formula of a compound of boron and hydrogen is \(\ce{BH_3}\). Its molar mass is \(27.7 \: \text{g/mol}\). Determine the molecular formula of the compound.

Solution Solutions to Example 6.9.1

Steps for Problem Solving	Determine the molecular formula of \(\ce{BH_3}\).
Identify the "given" information and what the problem is asking you to "find."	Given: Empirical formula \(= \ce{BH_3}\) Molar mass \(= 27.7 \: \text{g/mol}\) Find: Molecular formula \(= ?\)
Calculate the empirical formula mass (EFM).	\[\text{Empirical formula molar mass (EFM)} = 13.84 \: \text{g/mol} \nonumber \]
Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number.	\[\dfrac{\text{molar mass}}{\text{EFM}} = \dfrac{27.7 g/mol}{13.84 g/mol} = 2 \nonumber \]
Multiply all the subscripts in the empirical formula by the whole number found in step 2. The result is the molecular formula.	\[\ce{BH_3} \times 2 = \ce{B_2H_6} \nonumber \]
Write the molecular formula.	The molecular formula of the compound is \(\ce{B_2H_6}\).
Think about your result.	The molar mass of the molecular formula matches the molar mass of the compound.

Exercise \(\PageIndex{1}\)

Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What are the empirical and chemical formulas for ascorbic acid?

Answer Empirical FormulaC3H4O3Answer Molecular FormulaC6H8O6

Summary

A procedure is described that allows the calculation of the exact molecular formula for a compound.

Which compounds have empirical formula CH2O?

Thus empirical formula of the CH3COOH is CH2O.

Is C6H12O6 same as CH2O?

A molecular formula is a formula that represents the actual number of each atom in a compound. Whereas CH2O is the empirical formula for glucose, C6H12O6 is the molecular formula. An actual molecule of glucose contains six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.