 ### Note the Pattern

The molar fixed of any type of substance is its atom mass, molecular mass, or formula mass in grams per mole.

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The regular table list the atomic mass that carbon together 12.011 amu; the average molar mass of carbon—the fixed of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:

problem (formula) Atomic, Molecular, or Formula mass (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 different from the of carbon-12 and is no an integer since carbon occurs as a mixture the carbon-12, carbon-13, and carbon-14. One mole of carbon still has actually 6.022 × 1023 carbon atoms, but 98.89% that those atoms room carbon-12, 1.11% room carbon-13, and a map (about 1 atom in 1012) space carbon-14. (For more information, see section 1.6 ) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar massive of iodine is 126.90 g/mol. As soon as we attend to elements such together iodine and also sulfur, which happen as a diatomic molecule (I2) and also a polyatomic molecule (S8), respectively, molar massive usually refers to the fixed of 1 mol the atoms the the element—in this situation I and also S, not come the fixed of 1 mol that molecules of the facet (I2 and S8).

The molar mass of ethanol is the fixed of ethanol (C2H5OH) that includes 6.022 × 1023 ethanol molecules. As you calculated in example 1, the molecule mass the ethanol is 46.069 amu. Since 1 mol the ethanol includes 2 mol the carbon atoms (2 × 12.011 g), 6 mol that hydrogen atom (6 × 1.0079 g), and 1 mol that oxygen atoms (1 × 15.9994 g), that molar massive is 46.069 g/mol. Similarly, the formula mass of calcium phosphate is 310.177 amu, for this reason its molar massive is 310.177 g/mol. This is the fixed of calcium phosphate that includes 6.022 × 1023 formula units.

The mole is the communication of quantitative historicsweetsballroom.comistry. It gives historicsweetsballroom.comists v a way to transform easily between the mass of a substance and the variety of individual atoms, molecules, or formula devices of that substance. Conversely, it allows historicsweetsballroom.comists to calculate the fixed of a substance needed to acquire a desired number of atoms, molecules, or formula units. For example, to convert moles the a substance to mass, we usage the relationship

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

or, much more specifically,

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

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### Example 1.7.2

For 35.00 g the ethylene glycol (HOCH2CH2OH), i beg your pardon is provided in inks for ballpoint pens, calculate the number of

moles. Molecules.

Given: mass and also molecular formula

Asked for: number of mole and variety of molecules

Strategy:

A use the molecular formula of the link to calculate its molecular mass in grams per mole.

B transform from mass to mole by separating the mass given by the compound’s molar mass.

C transform from moles to molecule by multiplying the number of moles by Avogadro’s number.

Solution:

A The molecule mass the ethylene glycol deserve to be calculated indigenous its molecule formula utilizing the method illustrated in instance 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 fixed of ethanol = 62.068 amu

The molar massive of ethylene glycol is 62.068 g/mol

B The variety of moles of ethylene glycol current in 35.00 g have the right to be calculated by separating the fixed (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 estimate the prize before you carry out the yes, really calculation. In this case, the mass provided (35.00 g) is less 보다 the molar mass, for this reason the answer have to be less than 1 mol. The calculated prize (0.5639 mol) is certainly less 보다 1 mol, therefore we have probably not made a significant error in the calculations.

C To calculation the number of molecules in the sample, we multiply the variety of moles through 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 each compound.

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

Given: number the moles and also molecular or empirical formula

Strategy:

A calculation the molecular mass the the link in grams from its molecule formula (if covalent) or empirical formula (if ionic).

B transform from mole to mass by multiply the mole of the compound provided by its molar mass.

Solution:

We start by calculating the molecular mass the S2Cl2 and also the formula massive of Ca(ClO)2.

A The molar massive of S2Cl2 is acquired from its molecule mass together follows:

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

The molar fixed of S2Cl2 is 135.036 g/mol.

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The fixed of 1.75 mol that S2Cl2 is calculated as 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 massive of Ca(ClO)2 is derived 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 massive 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 massive of 0.0122 mol of each compound.

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