 ### Note the Pattern

The molar massive of any kind of substance is its atomic mass, molecular mass, or formula fixed in grams every mole.

You are watching: Calculate the mass percent of chlorine in ccl3 f (freon-11).

The regular table lists the atom mass of carbon as 12.011 amu; the typical molar massive of carbon—the massive of 6.022 × 1023 carbon atoms—is thus 12.011 g/mol:

problem (formula) Atomic, Molecular, or Formula massive (amu) Molar fixed (g/mol)
carbon (C) 12.011 (atomic mass) 12.011
ethanol (C2H5OH) 46.069 (molecular mass) 46.069
calcium phosphate 310.177 (formula mass) 310.177

The molar fixed of naturally developing carbon is various from that of carbon-12 and also is not an integer due to the fact that carbon occurs together a mixture of carbon-12, carbon-13, and also carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% that those atoms room carbon-12, 1.11% space carbon-13, and a trace (about 1 atom in 1012) room carbon-14. (For more information, see section 1.6 ) Similarly, the molar fixed of uranium is 238.03 g/mol, and the molar massive of iodine is 126.90 g/mol. As soon as we deal with elements such together iodine and sulfur, which take place as a diatomic molecule (I2) and also a polyatomic molecule (S8), respectively, molar fixed usually describes the mass of 1 mol that atoms of the element—in this situation I and S, not come the mass of 1 mol of molecules that the element (I2 and also S8).

The molar mass of ethanol is the mass of ethanol (C2H5OH) that contains 6.022 × 1023 ethanol molecules. Together you calculation in example 1, the molecular mass the ethanol is 46.069 amu. Since 1 mol the ethanol contains 2 mol the carbon atoms (2 × 12.011 g), 6 mol the hydrogen atoms (6 × 1.0079 g), and also 1 mol that oxygen atoms (1 × 15.9994 g), its molar massive is 46.069 g/mol. Similarly, the formula massive of calcium phosphate is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that consists of 6.022 × 1023 formula units.

The mole is the basis of quantitative dearteassociazione.orgistry. It provides dearteassociazione.orgists with a method to convert easily between the fixed of a substance and also the variety of individual atoms, molecules, or formula devices of the substance. Vice versa, it permits dearteassociazione.orgists to calculation the fixed of a substance essential to achieve a desired variety of atoms, molecules, or formula units. Because that example, to convert moles that a problem to mass, we use the relationship

$$(moles)(molar \; mass) \rightarrow mass \tag1.71$$

or, an ext specifically,

$$moles\left ( \dfracgramsmole \right ) = grams$$​

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### Example 1.7.2

For 35.00 g of ethylene glycol (HOCH2CH2OH), which is supplied in inks because that ballpoint pens, calculation the number of

moles. Molecules.

Given: mass and molecular formula

Asked for: number of mole and variety of molecules

Strategy:

A use the molecular formula that the link to calculation its molecular mass in grams per mole.

B convert from massive to mole by separating the mass provided by the compound’s molar mass.

C transform from mole to molecules by multiplying the variety of moles by Avogadro’s number.

Solution:

A The molecular mass of ethylene glycol have the right to be calculated indigenous its molecule formula using the technique illustrated in example 1:

 2C (2 atoms)(12.011 amu/atom) = 24.022 amu 6H (6 atoms)(1.0079 amu/atom) = 6.0474 amu 2O (2 atoms)(15.9994 amu/atom) = 31.9988 amu C2H6O molecular mass of ethanol = 62.068 amu

The molar massive of ethylene glycol is 62.068 g/mol

B The number of moles of ethylene glycol current in 35.00 g deserve to be calculated by dividing the massive (in grams) through the molar mass (in grams per mole):

$$\; 35.00\; g\; ethylene glycol\left ( \frac1\; mol\; ethylene\; glycol\; (g))62.068\; g\; ethylene\; glycol \right )=0.5639\; mol\; ethylene\; glycol$$

It is constantly a an excellent idea to calculation the prize before you execute the really calculation. In this case, the mass offered (35.00 g) is less 보다 the molar mass, so the answer need to be less than 1 mol. The calculated price (0.5639 mol) is indeed less than 1 mol, therefore we have probably not made a major error in the calculations.

C To calculate the number of molecules in the sample, us multiply the number of moles by Avogadro’s number:

$$molecules\; of\; ethylene\; glycol=0.5639\; mol\left ( \dfrac6.022\times 10^231\; mol \right )$$

$$= 3.396\times 10^23\; molecules​$$

Exercise

For 75.0 g that CCl3F (Freon-11), calculation the number of

moles. Molecules.

0.546 mol 3.29 × 1023 molecules

### Example 1.7.3

Calculate the massive of 1.75 mol of every compound.

S2Cl2 (common name: sulfur monochloride; organized name: disulfur dichloride) Ca(ClO)2 (calcium hypochlorite)

Given: number that moles and molecular or empirical formula

Strategy:

A calculate the molecule mass that the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).

B convert from mole to massive by multiplying the mole of the compound given by its molar mass.

Solution:

We begin by calculating the molecule mass the S2Cl2 and also the formula mass of Ca(ClO)2.

A The molar massive of S2Cl2 is derived from its molecule mass as follows:

 2S (2 atoms)(32.065 amu/atom) = 64.130 amu 2Cl (2 atoms)(35.353 amu/atom) = 70.906 amu S2Cl​2 molecular mass of S2Cl​2 = 135.036 amu

The molar massive of S2Cl2 is 135.036 g/mol.

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The mass of 1.75 mol the S2Cl2 is calculated together follows:

$$moles\; S_2Cl_2 \left = mass\; S_2Cl_2$$

$$1.75\; mol\; S_2Cl_2\left ( \dfrac135.036\; g\; S_2Cl_21\;mol\;S_2Cl_2 \right )=236\;g\; S_2Cl_2$$

B The formula fixed of Ca(ClO)2 is obtained as follows:

 1Ca (1 atom )(40.078 amu/atom) = 40.078 amu 2Cl (2 atoms)(35.453 amu/atom) = 70.906 amu 2O (2 atoms)(15.9994 amu/atom) = 31.9988 amu Ca(ClO)2 formula fixed of Ca(ClO)2 = 142.983 amu

The molar fixed of Ca(ClO)2 is 142.983 g/mol

$$moles\; Ca\left ( ClO \right )_2\left < \dfracmolar\; mass\; Ca\left ( ClO \right )_21\; mol\; Ca\left ( ClO \right )_2 \right >=mass\; Ca\left ( ClO \right )_2$$

$$1.75\; mol\; Ca\left ( ClO \right )_2\left < \dfrac142.983\; g Ca\left ( ClO \right )_21\; mol\; Ca\left ( ClO \right )_2 \right >=250.\; g\; Ca\left ( ClO \right )_2$$​

Exercise

Calculate the fixed of 0.0122 mol of each compound.

Si3N4 (silicon nitride), used as bearings and rollers (CH3)3N (trimethylamine), a corrosion inhibitor