identify the molar mass because that a compound or molecule. Convert from mole to grams and grams to moles.

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In the previous section we identified molar mass together the fixed of one mole of anything, or the massive of 6.022 x 1023 of the thing. In this section we"re going come look at just how this used to molecule or compounds.

Molar Mass

Molar mass is defined as the massive of one mole that representative particles of a substance. By looking in ~ a periodic table, we can conclude that the molar massive of lithium is (6.94 : extg), the molar massive of zinc is (65.38 : extg), and the molar mass of yellow is (196.97 : extg). Each of this quantities has (6.02 imes 10^23) atoms of that certain element. The devices for molar mass are grams per mole or ( extg/mol).

Molar Masses that Compounds

The molecule formula that the compound carbon dioxide is (ceCO_2). One molecule that carbon dioxide is composed of 1 atom that carbon and 2 atoms of oxygen. We deserve to calculate the massive of one molecule that carbon dioxide by including together the masses that 1 atom the carbon and also 2 atom of oxygen.

<12.01 : extamu + 2 left( 16.00 : extamu ight) = 44.01 : extamu>

The molecular mass the a link is the massive of one molecule of that compound. The molecule mass of carbon dioxide is (44.01 : extamu).

The molar fixed of any type of compound is the massive in grams the one mole of that compound. One mole the carbon dioxide molecules has a massive of (44.01 : extg), while one mole of sodium sulfide formula units has actually a fixed of (78.04 : extg). The molar masses room (44.01 : extg/mol) and (78.04 : extg/mol) respectively. In both cases, that is the massive of (6.02 imes 10^23) representative particles. The representative particle of (ceCO_2) is the molecule, while because that (ceNa_2S) that is the formula unit.

Example (PageIndex1)

Calcium nitrate, (ceCa(NO_3)_2), is used as a component in fertilizer. Identify the molar fixed of calcium nitrate.


Step 1: list the known and also unknown quantities and also plan the problem.


Formula (= ceCa(NO_3)_2) Molar fixed (ceCa = 40.08 : extg/mol) Molar mass (ceN = 14.01 : extg/mol) Molar fixed (ceO = 16.00 : extg/mol)


Molar massive (ceCa(NO_3)_2)

First we must analyze the formula. Because the (ceCa) lacks a subscript, there is one (ceCa) atom per formula unit. The 2 external the parentheses means that there room two nitrate ions per formula unit and each nitrate ion is composed of one nitrogen atom and three oxygen atoms every formula unit. Thus, (1 : extmol) the calcium nitrate consists of (1 : extmol) that (ceCa) atoms, (2 : extmol) of (ceN) atoms, and also (6 : extmol) the (ceO) atoms.

Step 2: Calculate

Use the molar masses of every atom along with the variety of atoms in the formula and include together.

<1 : extmol : ceCa imes frac40.08 : extg : ceCa1 : extmol : ceCa = 40.08 : extg : ceCa>

<2 : extmol : ceN imes frac14.01 : extg : ceN1 : extmol : ceN = 28.02 : extg : ceN>

<6 : extmol : ceO imes frac16.00 : extg : ceO1 : extmol : ceO = 96.00 : extg : ceO>

Molar massive of (ceCa(NO_3)_2 = 40.08 : extg + 28.02 : extg + 96.00 : extg = 164.10 : extg/mol)

Here room some additional examples:

The massive of a hydrogen atom is 1.0079 amu; the fixed of 1 mol the hydrogen atoms is 1.0079 g. Element hydrogen exists as a diatomic molecule, H2. One molecule has a fixed of 1.0079 + 1.0079 = 2.0158 amu, if 1 mol H2 has actually a mass of 2.0158 g. A molecule of H2O has actually a massive of about 18.01 amu; 1 mol H2O has a fixed of 18.01 g. A single unit of NaCl has a fixed of 58.45 amu; NaCl has actually a molar mass of 58.45 g.

In every of these moles of substances, there are 6.022 × 1023 units:

6.022 × 1023 atoms of H 6.022 × 1023 molecules of H2 and H2O, 6.022 × 1023 devices of NaCl ions.

These relationships offer us plenty of avenues to construct conversion determinants for straightforward calculations.

Example (PageIndex2)

What is the molar mass of C6H12O6?


To determine the molar mass, we simply add the atom masses the the atoms in the molecule formula however express the complete in grams every mole, no atomic mass units. The masses of the atoms deserve to be taken from the periodic table.

6 C = 6 × 12.011 = 72.066
12 H = 12 × 1.0079 = 12.0948
6 O = 6 × 15.999 = 95.994
TOTAL = 180.155 g/mol

Per convention, the unit grams every mole is created as a fraction.

Exercise (PageIndex2)

What is the molar fixed of AgNO3?


169.87 g/mol

Knowing the molar fixed of a substance, we have the right to calculate the variety of moles in a particular mass the a substance and vice versa, as these examples illustrate. The molar fixed is supplied as the conversion factor.

Example (PageIndex3)

What is the fixed of 3.56 mol of HgCl2? The molar fixed of HgCl2 is 271.49 g/mol.


Use the molar mass together a counter factor between moles and also grams. Since we want to publication the mole unit and introduce the gram unit, we have the right to use the molar mass together given:

<3.56, cancelmol, HgCl_2 imes frac271.49, g, HgCl_2cancelmol, HgCl_2=967, g, HgCl_2>

Exercise (PageIndex3)

What is the mass of 33.7 mol the H2O?


607 g

Example (PageIndex4)

How many moles of H2O are existing in 240.0 g that water (about the mass of a cup that water)?


Use the molar mass of H2O as a conversion factor from mass to moles. The molar fixed of water is (1.0079 + 1.0079 + 15.999) = 18.015 g/mol. However, since we desire to cancel the gram unit and introduce moles, we should take the mutual of this quantity, or 1 mol/18.015 g:

<240.0, cancelg, H_2O imes frac1, mol, H_2O18.015cancelg, H_2O=13.32, mol, H_2O>

Exercise (PageIndex4)

How many moles are present in 35.6 g of H2SO4 (molar fixed = 98.08 g/mol)?


0.363 mol

Other conversion components can be merged with the an interpretation of mole-density, for example.

Example (PageIndex5)

The thickness of ethanol is 0.789 g/mL. How numerous moles room in 100.0 mL the ethanol? The molar massive of ethanol is 46.08 g/mol.


Here, us use density to convert from volume come mass and also then use the molar fixed to determine the number of moles.

<100cancelml: ethanol imes frac0.789, gcancelml imes frac1, mol46.08, cancelg=1.71, mol, ethanol>

Exercise (PageIndex5)

If the thickness of benzene, C6H6, is 0.879 g/mL, how numerous moles are current in 17.9 mL that benzene?


0.201 mol

Converting in between mass, mole or atom of a compound to mass, moles or atom of that elements

Now that we know just how to convert from atom to molecules, from molecules to moles and also from moles to grams we have the right to string this conversion components together to settle more complex problems.

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Example (PageIndex6): Converting in between grams and also Atoms

How plenty of atoms the hydrogen are in 4.6 g the CH3OH?


First we require to determine the fixed of one mole of methane (CH3OH).

Using the routine table to find the mass for each mole the our facets we have:

<1, mole, C ,= 1, cancelmole, C, imes left(frac12.011, g, C1,cancelmole,C ight), = 12.011 , g, C>

<4, mole, H ,= 4, cancelmole, H, imes left(frac1.008, g, H1,cancelmole,H ight), = 4.032 ,g, H>

<1, mole, O ,= 1, cancelmole, O, imes left(frac15.999, g, O1,cancelmole,O ight), = 15.999 , g,O>

Adding the masses of our individual facets have:

<12.011, g ,+ ,4.032,g, +, 15.999, g, =,32.042, g, CH_3OH>

As we were calculating the molar massive of CH3OH we have

<32.042, g, CH_3OH, = ,1, mole ,CH_3OH>

Which we can use together a conversion factor

We likewise know the for every molecule of CH3OH we have actually 4 atom of H.

Now we have the right to go earlier to the starting value offered in the question:

<4.6,cancelg, CH_3OH, imes,left(frac1, cancelmole, CH_3OH32.042,cancelg,CH_3OH ight) imesleft(frac6.022, x, 10^23,cancel,molecules, CH_3OH1 ,cancelmole, CH_3OH ight) imesleft(frac4, atoms, H1 cancelmolecule ,CH_3OH ight) = 3.5,x,10^23, atoms, H>

Exercise (PageIndex6)

How plenty of atoms the H are there in 2.06 grams of (ceH_2O)?


1.38 x 1023 atom H

Example (PageIndex7): converting from Grams come grams

How plenty of grams that oxygen are in 3.45 g of H3PO4


First we need to determine the mass of one mole the phosphoric acid (ceH_3PO_4)

We have:

<3, mole, H ,= 3, cancelmole, H, imes left(frac1.008, g, H1,cancelmole,H ight), = 3.024,g, H>

<1, mole, p ,= 1, cancelmole, P, imes left(frac30.974, g, P1,cancelmole,P ight), = 30.974 ,g, P>

<4, mole, O ,= 4, cancelmole, O, imes left(frac15.999, g, O1,cancelmole,O ight), = 63.996 ,g, O>

Adding the masses of ours individual aspects have:

<3.024, g ,+ ,30.974,g, +, 63.996, g, =,97.994, g, H_3PO_4>

As we were calculating the molar fixed of CH3OH we have

<97.994, g, H_3PO_4, = ,1, mole ,H_3PO_4>

Which we can use as a conversion factor

To calculate the molar massive of (ceH_3PO_4) we provided the idea the we have 3 moles of H, 1 mole the P and 4 moles of O for every one mole the (ceH_3PO_4). We recall the we can make our very own conversion factors as lengthy as the top and bottom space equal to every other. So lot the same method I can give you four quarter or 1 dollar, if ns hand girlfriend a mole the (ceH_3PO_4) I"ve handed girlfriend 3 mole the H, 1 mole of P and also 4 mole of O. Together this is true over there is an additional set that conversion factors we can use:

Now we have the right to go earlier to the starting value provided in the question:

<3.45,cancelg,H_3PO_4, imes,left(frac1,cancelmole,H_3PO_497.994,cancelg,H_3PO_4 ight) imesleft(frac4,cancelmole,O1,cancel,mole, H_3PO_4 ight) imesleft(frac15.999, g, O1 cancelmole ,O ight) = 2.25, g, O>

Exercise (PageIndex6)

How many grams the H space there in 3.45 grams the (ceH_3PO_4)?


0.106 grams H

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