Molecular mass

Calculation of relative molecular mass

Since the total mass is conserved with high precision when atoms combine into a single structure, the relative molecular mass is found by simple addition. To do this, one must take the relative atomic masses of each chemical element making up the compound and multiply them by the number of corresponding atoms (subscripts in the chemical formula). Mandatory rule: tabular values of atomic masses are rounded to a whole number during these calculations.

In the form of a classical mathematical expression, the algorithm looks like this: $M_r = \sum N_i A_{ri}$, where the variable $N_i$ denotes the number of atoms of a specific element, and $A_{ri}$ is its relative atomic mass.

Let us consider the calculation using the classic example of sulfuric acid ($H_2SO_4$). A single molecule of it contains two hydrogen atoms, one sulfur atom, and four oxygen atoms. We add up their masses taking into account the subscripts in the chemical formula:

• $M_r(H_2SO_4) = 2 \cdot 1 + 1 \cdot 32 + 4 \cdot 16$.

After performing the arithmetic operations, we obtain the final value — 98.

Mass fractions of elements in a compound

Any chemical compound is characterized by a strictly defined composition, which in chemistry is usually expressed through mass fractions (denoted by the Greek letter $\omega$). The mass fraction clearly demonstrates what exact part of the total mass of the entire molecule is made up by the total mass of a specific element.

To find this indicator, it is necessary to divide the mass of all atoms of the target element in the molecule by the total molecular mass of the entire substance ($M_r$). The formula looks like this: $\omega_i = (A_i N_{ri}) / \Sigma A_i N_{ri}$.

For example, let us calculate the mass fractions of elements for a glucose molecule, which has the formula $C_6H_{12}O_6$. First, we find the total molecular mass: we add $6 \cdot 12$ for carbon atoms, $12 \cdot 1$ for hydrogen atoms, and $6 \cdot 16$ for oxygen atoms. The final $M_r$ is 180.

Now let us successively calculate the mass fractions:

• Carbon (C): we divide 72 by 180, which results in 0.40 (or 40%).
• Hydrogen (H): we divide 12 by 180, getting 0.067 (or 6.7%).
• Oxygen (O): we divide 96 by 180, calculating a value of 0.533 (or 53.3%).

Determining the chemical formula from composition

Knowledge of the percentage elemental composition makes it possible to solve an important inverse problem — determining the chemical formula of an unknown substance. In such calculation problems, the mass fraction of an element in percent is conventionally taken as the mass of this element in 100 mass parts (m.p.) of the analyzed sample.

Suppose that during the analysis it turned out that the chemical compound consists exclusively of carbon and sulfur, with the mass fraction of carbon being 15.8%. Consequently, the remaining $100 - 15.8 = 84.2$ m.p. are accounted for by sulfur.

Next, to determine the number of atoms, a mathematical proportion is set up. An unknown mass of sulfur corresponds to one carbon atom, whose atomic mass is 12 a.m.u. This mass is easy to calculate: $(12 \cdot 84.2) / 15.8$. The result is 64 a.m.u.

Since the atomic mass of a single sulfur atom is 32, we divide 64 by 32 and get exactly two sulfur atoms. Thus, based on the calculations, the exact chemical formula of the substance is established — $CS_2$.

Number of molecules and types of chemical formulas

If the mass of a substance in grams is known in a problem, the exact number of molecules in a given volume can be calculated. For this, a standard physical conversion factor is used: 1 a.m.u. is equivalent to $1.66 \cdot 10^{-24}$ g. For example, to find out the number of molecules in 5 grams of sulfuric acid, it is necessary to divide the mass of the sample by the weight of a single molecule in grams: $5 / (98 \cdot 1.66 \cdot 10^{-24})$. The final calculation result will be $3.07 \cdot 10^{22}$ molecules.

In addition, when analyzing the structure of substances, chemists often have to distinguish between the empirical (simplest) and true (molecular) formulas. For example, an analysis of the composition of ethane gas shows that it consists of 80% carbon and 20% hydrogen by mass. If this ratio is converted into atomic mass units, it turns out that 12 a.m.u. of carbon correspond to 3 a.m.u. of hydrogen (i.e., three atoms).

Such a calculation yields the empirical formula $CH_3$, which merely reflects the minimum proportion of elements. However, in chemical calculations, the real molecular formula — $C_2H_6$ — is used, which fully describes the true structure of the molecule.

Class and exam questions

What else to study on this topic

• Calculation of the mass of an individual molecule of a substance in grams.
• Differences in the use of the concept of molecular mass for molecules and formula units.
• Determination of the true molecular formula of organic gases by their density.
• Algorithms for calculating the mass fraction of chemical elements in complex polymer compounds.

Other topics in the section "Chapter 1. Substances. Properties and Structure"

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