Determine the molar mass for a compound or molecule. Convert from mole to grams and grams come moles.

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In the previous section we identified molar mass as the fixed of one mole of anything, or the massive of 6.022 x 1023 of the thing. In this ar we"re going come look at exactly how this applied to molecules or compounds.


Molar Mass

Molar mass is defined as the mass of one mole the representative corpuscle of a substance. Through looking in ~ a periodic table, we have the right to conclude the the molar fixed of lithium is \(6.94 \: \textg\), the molar mass of zinc is \(65.38 \: \textg\), and also the molar massive of gold is \(196.97 \: \textg\). Each of this quantities includes \(6.02 \times 10^23\) atoms of that certain element. The systems for molar mass are grams per mole or \(\textg/mol\).


Molar Masses that Compounds

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

\<12.01 \: \textamu + 2 \left( 16.00 \: \textamu \right) = 44.01 \: \textamu\>

The molecular mass the a link is the fixed of one molecule of the compound. The molecule mass the carbon dioxide is \(44.01 \: \textamu\).

The molar massive of any compound is the mass in grams that one mole of the compound. One mole that carbon dioxide molecules has actually a fixed of \(44.01 \: \textg\), when one mole of sodium sulfide formula units has actually a mass of \(78.04 \: \textg\). The molar masses are \(44.01 \: \textg/mol\) and \(78.04 \: \textg/mol\) respectively. In both cases, the is the massive of \(6.02 \times 10^23\) representative particles. The representative fragment of \(\ceCO_2\) is the molecule, while for \(\ceNa_2S\) it is the formula unit.


Example \(\PageIndex1\)

Calcium nitrate, \(\ceCa(NO_3)_2\), is provided as a ingredient in fertilizer. Determine the molar fixed of calcium nitrate.

Solution:

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

Known

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

Unknown

Molar fixed \(\ceCa(NO_3)_2\)

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

Step 2: Calculate

Use the molar masses of every atom together with the number of atoms in the formula and add together.

\<1 \: \textmol \: \ceCa \times \frac40.08 \: \textg \: \ceCa1 \: \textmol \: \ceCa = 40.08 \: \textg \: \ceCa\>

\<2 \: \textmol \: \ceN \times \frac14.01 \: \textg \: \ceN1 \: \textmol \: \ceN = 28.02 \: \textg \: \ceN\>

\<6 \: \textmol \: \ceO \times \frac16.00 \: \textg \: \ceO1 \: \textmol \: \ceO = 96.00 \: \textg \: \ceO\>

Molar mass of \(\ceCa(NO_3)_2 = 40.08 \: \textg + 28.02 \: \textg + 96.00 \: \textg = 164.10 \: \textg/mol\)


Here are some further examples:

The massive of a hydrogen atom is 1.0079 amu; the mass of 1 mol the hydrogen atom 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 a fixed of 2.0158 g. A molecule the H2O has a mass of about 18.01 amu; 1 mol H2O has a massive of 18.01 g. A single unit the NaCl has actually a massive of 58.45 amu; NaCl has actually a molar fixed of 58.45 g.

In each of this moles of substances, there space 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 simple calculations.


Example \(\PageIndex2\)

What is the molar massive of C6H12O6?

Solution

To identify the molar mass, us simply include the atom masses of the atom in the molecular formula but express the full in grams per mole, no atomic massive units. The masses that the atoms have the right 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 composed as a fraction.


Exercise \(\PageIndex2\)

What is the molar massive of AgNO3?

Answer

169.87 g/mol


Knowing the molar mass of a substance, we have the right to calculate the number of moles in a details mass the a substance and also vice versa, as these instances illustrate. The molar massive is supplied as the conversion factor.


Example \(\PageIndex3\)

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

Solution

Use the molar mass as a switch factor in between moles and also grams. Due to the fact that we want to publication the mole unit and also introduce the gram unit, we have the right to use the molar mass as given:

\<3.56\, \cancelmol\, HgCl_2\times \frac271.49\, g\, HgCl_2\cancelmol\, HgCl_2=967\, g\, HgCl_2\>


Exercise \(\PageIndex3\)

What is the massive of 33.7 mol of H2O?

Answer

607 g


Example \(\PageIndex4\)

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

Solution

Use the molar fixed 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, because we want to cancel the gram unit and also introduce moles, we have to take the mutual of this quantity, or 1 mol/18.015 g:

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


Exercise \(\PageIndex4\)

How countless moles are current in 35.6 g of H2SO4 (molar mass = 98.08 g/mol)?

Answer

0.363 mol


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


Example \(\PageIndex5\)

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

Solution

Here, we use density to convert from volume to mass and then usage the molar fixed to determine the variety of moles.

\<100\cancelml\: ethanol\times \frac0.789\, g\cancelml\times \frac1\, mol46.08\, \cancelg=1.71\, mol\, ethanol\>


Exercise \(\PageIndex5\)

If the density of benzene, C6H6, is 0.879 g/mL, how countless moles are current in 17.9 mL of benzene?

Answer

0.201 mol


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

Now that us know how to transform from atom to molecules, from molecule to moles and from moles to grams we can string these conversion determinants together to solve more complex problems.

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

How many atoms of hydrogen space in 4.6 g the CH3OH?

Solution

First we need to recognize the massive of one mole of methane (CH3OH).

Using the regular table to discover the mass because that each mole the our aspects we have:

\<1\, mole\, C \,= 1\, \cancelmole\, C\,\times \left(\frac12.011\, g\, C1\,\cancelmole\,C \right)\, = 12.011 \, g\, C\>

\<4\, mole\, H \,= 4\, \cancelmole\, H\,\times \left(\frac1.008\, g\, H1\,\cancelmole\,H \right)\, = 4.032 \,g\, H\>

\<1\, mole\, O \,= 1\, \cancelmole\, O\,\times \left(\frac15.999\, g\, O1\,\cancelmole\,O \right)\, = 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 mass of CH3OH we have

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

Which we can use together a counter factor

\<\frac32.042\, g\, CH_3OH1\, mole\, CH_3OH\, or\, \frac1\, mole\, CH_3OH32.042\, g\, CH_3OH\>

We additionally know that for every molecule of CH3OH we have 4 atoms of H.

Now we can go back to the beginning value given in the question:

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


Exercise \(\PageIndex6\)

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

Answer

1.38 x 1023 atom H


Example \(\PageIndex7\): converting from Grams to grams

How numerous grams that oxygen space in 3.45 g of H3PO4

Solution

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

We have:

\<3\, mole\, H \,= 3\, \cancelmole\, H\,\times \left(\frac1.008\, g\, H1\,\cancelmole\,H \right)\, = 3.024\,g\, H\>

\<1\, mole\, ns \,= 1\, \cancelmole\, P\,\times \left(\frac30.974\, g\, P1\,\cancelmole\,P \right)\, = 30.974 \,g\, P\>

\<4\, mole\, O \,= 4\, \cancelmole\, O\,\times \left(\frac15.999\, g\, O1\,\cancelmole\,O \right)\, = 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 massive of CH3OH we have

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

Which we can use together a switch factor

\<\frac97.994\, g\, H_3PO_41\, mole\, H_3PO_4\, or\, \frac1\, mole\, H_3PO_497.994\, g\, H_3PO_4\>

To calculate the molar mass of \(\ceH_3PO_4)\ we provided the idea the we have 3 mole of H, 1 mole of P and also 4 mole of O because that every one mole of \(\ceH_3PO_4). Us recall the we deserve to make our very own conversion determinants as long as the top and bottom room equal to every other. So lot the same way I can offer you 4 quarter or 1 dollar, if ns hand you a mole the \(\ceH_3PO_4) I"ve handed friend 3 mole of H, 1 mole that P and also 4 moles of O. Together this is true there is second set of conversion factors we can use:

\<\left(\frac3\, moles\, H1\, mole\, H_3PO_4\right)\, and also \, \left(\frac1\, mole\, P1\, mole\, H_3PO_4\right)\, and\, \left(\frac4\,moles\, O1\, mole\, H_3PO_4\right)\>

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

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


Exercise \(\PageIndex6\)

How countless grams that H space there in 3.45 grams of \(\ceH_3PO_4\)?

Answer

0.106 grams H


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