Show Let's try to create a conversion factor that will take us directly from the number of grams of calcium oxide to the number of formula units present in the sample. For starters, look up the molar mass of calcium oxide
This tells you that #1# mole of calcium oxide has a mass of #"56.0774 g"#, which can be written as a conversion factor
Now, you know that in order to have #1# mole of calcium oxide, you need to have #6.022 * 10^(23)# formula units of calcium oxide #-># this is known as Avogadro's constant. This can be written as a conversion factor as well
As you can see, the two conversion factors have a common quantity. If you multiply these two conversion factors
you will get
You now have a conversion factor that takes you from grams to formula units or vice versa. So, you know that your sample has a mass of #"0.335 g"#. Multiply this by the conversion factor to get
The answer is rounded to three sig figs, the number of sig figs you have for the mass of calcium oxide. There are #3.60xx10^21color(white)(.)"formula units"# in #0.335 g CaO"#. There are #6.022xx10^23# in 1 mole of anything, including formula units. You need to determine the number of moles in #"0.335 g CaO"#. Once you know the number of moles of #"CaO"#, you can determine the number of formula units by multiplying the number of moles by #6.022xx10^23#. #color(blue)("Number of Moles"# You need to #color(red)("determine the molar mass"# of #"CaO"#, which is the sum of the atomic weights of each element on the periodic table in grams/mole, or g/mol. #"Ca:"##"40.078 g/mol"# #"Molar mass CaO"="40.078 g/mol + 15.999 g/mol"="56.077 g/mol"# #color(red)("Determine the number of moles"# by multiplying the given mass of #"CaO"# by the inverse of its molar mass. #0.335color(red)cancel(color(black)("g CaO"))xx(1"mol CaO")/(56.077color(red)cancel(color(black)("g CaO")))="0.00597 mol CaO"# #color(blue)("Number of Formula Units"# Multiply the mol #"CaO"# by #6.022xx10^23# formula units/mol. #0.00597color(red)cancel(color(black)("mol CaO"))xx(6.022xx10^23"formula units CaO")/(1color(red)cancel(color(black)("mol CaO")))=3.60xx10^21color(white)(.)"formula units"# rounded to three significant figures First you need to determine the number of moles in #"0.67 g CaO"#. Then you will multiply the number of moles by #6.022xx10^23 "formula units/mol"#. To determine the molar mass of a compound, add the atomic weight on the periodic table in g/mol times each element's subscript. Since the formula unit #"CaO"# has no subscripts, they are understood to be #1#. Molar Masses of Calcium and Oxygen #"Ca:"## "40.078 g/mol"# Molar Mass of Calcium Oxide #("40.078 g/mol"xx1) + ("15.999 g/mol"xx1)="56.077 g/mol"# Determining Moles CaO #(0.67cancel"g" "CaO")/(56.077cancel"g"/"mol" "CaO")="0.01195 mol CaO"# I'm keeping some extra digits in order to reduce rounding errors. The final answer will be rounded to two significant figures. Determine Formula Units of
CaO #0.01195cancel"mol"xx(6.022xx10^23 "formula units")/cancel"mol"=7.2xx10^22 "formula units CaO"# Skills to Develop Figure \(\PageIndex{1}\) shows that we need 2 hydrogen atoms and 1 oxygen atom to make 1 water molecule. If we want to make 2 water molecules, we will need 4 hydrogen atoms and 2 oxygen atoms. If we want to make 5 molecules of water, we need 10 hydrogen atoms and 5 oxygen atoms. The ratio of atoms we will need to make any number of water molecules is the same: 2 hydrogen atoms to 1 oxygen atom. Figure \(\PageIndex{1}\) Water Molecules. The ratio of hydrogen atoms to oxygen atoms used to make water molecules is always 2:1, no matter how many water molecules are being made. One problem we have, however, is that it is extremely difficult, if not impossible, to organize atoms one at a time. As stated in the introduction, we deal with billions of atoms at a time. How can we keep track of so many atoms (and molecules) at a time? We do it by using mass rather than by counting individual atoms. A hydrogen atom has a mass of approximately 1 u. An oxygen atom has a mass of approximately 16 u. The ratio of the mass of an oxygen atom to the mass of a hydrogen atom is therefore approximately 16:1. If we have 2 atoms of each element, the ratio of their masses is approximately 32:2, which reduces to 16:1—the same ratio. If we have 12 atoms of each element, the ratio of their total masses is approximately (12 × 16):(12 × 1), or 192:12, which also reduces to 16:1. If we have 100 atoms of each element, the ratio of the masses is approximately 1,600:100, which again reduces to 16:1. As long as we have equal numbers of hydrogen and oxygen atoms, the ratio of the masses will always be 16:1. The same consistency is seen when ratios of the masses of other elements are compared. For example, the ratio of the masses of silicon atoms to equal numbers of hydrogen atoms is always approximately 28:1, while the ratio of the masses of calcium atoms to equal numbers of lithium atoms is approximately 40:7. So we have established that the masses of atoms are constant with respect to each other, as long as we have the same number of each type of atom. Consider a more macroscopic example. If a sample contains 40 g of Ca, this sample has the same number of atoms as there are in a sample of 7 g of Li. What we need, then, is a number that represents a convenient quantity of atoms so we can relate macroscopic quantities of substances. Clearly even 12 atoms are too few because atoms themselves are so small. We need a number that represents billions and billions of atoms. Chemists use the term mole to represent a large number of atoms or molecules. Just as a dozen implies 12 things, a mole (mol) represents 6.022 × 1023 things. The number 6.022 × 1023, called Avogadro’s number after the 19th-century chemist Amedeo Avogadro, is the number we use in chemistry to represent macroscopic amounts of atoms and molecules. Thus, if we have 6.022 × 1023 O atoms, we say we have 1 mol of O atoms. If we have 2 mol of Na atoms, we have 2 × (6.022 × 1023) Na atoms, or 1.2044 × 1024 Na atoms. Similarly, if we have 0.5 mol of benzene (C6H6) molecules, we have 0.5 × (6.022 × 1023) C6H6 molecules, or 3.011 × 1023 C6H6 molecules. Note A mole represents a very large number! If 1 mol of quarters were stacked in a column, it could stretch back and forth between Earth and the sun 6.8 billion times. Notice that we are applying the mole unit to different types of chemical entities. In these examples, we cited moles of atoms and moles of molecules. The word mole represents a number of things—6.022 × 1023 of them—but does not by itself specify what “they” are. They can be atoms, formula units (of ionic compounds), or molecules. That information still needs to be specified. Because 1 H2 molecule contains 2 H atoms, 1 mol of H2 molecules (6.022 × 1023 molecules) has 2 mol of H atoms. Using formulas to indicate how many atoms of each element we have in a substance, we can relate the number of moles of molecules to the number of moles of atoms. For example, in 1 mol of ethanol (C2H6O), we can construct the following relationships (Table \(\PageIndex{1}\)): Table \(\PageIndex{1}\): Molecular Relationships
The following example illustrates how we can use these relationships as conversion factors. Example \(\PageIndex{1}\) If a sample consists of 2.5 mol of ethanol (C2H6O), how many moles of carbon atoms, hydrogen atoms, and oxygen atoms does it have? SOLUTION Using the relationships in Table \(\PageIndex{1}\), we apply the appropriate conversion factor for each element: Note how the unit mol C2H6O molecules cancels algebraically. Similar equations can be constructed for determining the number of H and O atoms: \(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{6\: mol\: H\: atoms}{1\: mol\: C_2H_6O\: molecules}=15\: mol\: H\: atoms}\) \(\mathrm{2.5\: mol\: C_2H_6O\: molecules\times\dfrac{1\: mol\: O\: atoms}{1\: mol\: C_2H_6O\: molecules}=2.5\: mol\: O\: atoms}\) Exercise \(\PageIndex{1}\) If a sample contains 6.75 mol of Na2SO4, how many moles of sodium atoms, sulfur atoms, and oxygen atoms does it have? The fact that 1 mol equals 6.022 × 1023 items can also be used as a conversion factor. Example \(\PageIndex{2}\) How many formula units are present in 2.34 mol of NaCl? How many ions are in 2.34 mol? SOLUTION Typically in a problem like this, we start with what we are given and apply the appropriate conversion factor. Here, we are given a quantity of 2.34 mol of NaCl, to which we can apply the definition of a mole as a conversion factor: \(\mathrm{2.34\: mol\: NaCl\times\dfrac{6.022\times10^{23}\: NaCl\: units}{1\: mol\: NaCl}=1.41\times10^{24}\: NaCl\: units}\) Because there are two ions per formula unit, there are \(\mathrm{1.41\times10^{24}\: NaCl\: units\times\dfrac{2\: ions}{NaCl\: units}=2.82\times10^{24}\: ions}\) in the sample. Exercise \(\PageIndex{2}\) How many molecules are present in 16.02 mol of C4H10? How many atoms are in 16.02 mol? Concept Review Exercise
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How do you convert formula units to moles?To convert formula units to moles you divide the quantity of formula units by Avogadro's number, which is 6.022 x 1023 formula units/mole.
How do you find the number of formula units in moles?The unit is denoted by mol.. The formula for the number of moles formula is expressed as.. Given.. Number of moles formula is.. Number of moles = Mass of substance / Mass of one mole.. Number of moles = 95 / 86.94.. What is the number of formula units?The formula units of any compound are determined by multiplying the moles of the given compound by the Avogadro number i.e., 6.023×1023 6.023 × 10 23 .
What is formula unit in mole concept?This formula can be written as: Number of Atoms or Molecules = (Number of Moles)*(6.022*1023) The relationship between the atomic mass unit (amu) and the gram is given by: 1 amu = (1gram)/(6.022*1023) = 1.66*10-24 grams. Therefore, the mass of one mole of an element will be equal to its atomic mass in grams.
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