Why Significant Digits?

Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement.  A number with more significant digits is more precise.  For example, 8.00 cm is more precise than 8.0 cm.

Recording Measurements

When recording a measurement we include all of the known digits plus a final estimated digit.  It’s sometimes easier to think of this in terms of recording all of the known “places” (ones, tenths, hundredths) plus a final estimated place.  For example, in the first ruler below it is marked every one centimeter so we know the ones place and could record 2.  But we must also estimate one, and only one, additional digit.  So in this case we might record 2.3 cm or perhaps 2.4 cm. Either would be correct as the 2 (ones place) is precisely known while the final digit (tenths place) is estimated.

Now consider the next ruler.  In this case, the ruler is marked in both ones and tenths meaning that we can clearly see the first part of the measurement is 2.3. However, we must again estimate one additional digit or place - 2.35 cm.

Counting Significant Digits

If a measurement has already been made and provided, then we can determine how many significant digits are present by following a few simple rules.

1. All non-zero numbers are significant.

For example, the number 843 would have three significant digits are they are all non-zero.

2. Zeros between non-zero numbers are significant.

The number 307 would also have three significant digits as the zero is sandwiched between the non-zero numbers 3 and 7.

3. Leading zeros before a number are not significant.

In the number 0.0025, the leading zeros (those to the left of the non-zero numbers) are not significant.  Therefore, only the 2 & 5 are counted meaning it has two significant digits.  The leading zeros are known as placeholder zeros as they do not add to the precision of the measurement, they simply occupy the ones, tenths, and hundredths places. This will become more clear in our lesson on Scientific Notation as 0.0025 could also be written as 2.5×10^-3, completely eliminating the leading zeros.

4. Trailing zeros after a number are not significant unless there’s a decimal point.

Consider three different measurements: 250 versus 250. versus 250.0

250 – The trailing zeros (those to the right of the non-zero numbers) are also placeholders and thus do not add to the precision of the measurement. Thus, there are only two significant digits from the 2 & 5.  Again, as we’ll see in our lesson on Scientific Notation, this could be written 2.5×10² completely eliminating the final placeholder zero.

250. – The decimal point indicates the measurement is precisely 250. not around 250 as in the previous example.  This number would have three significant digits.

250.0 – Again, the decimal point indicates that the trailing zeros are significant and should be counted meaning there are four significant digits.

Counting Significant Digits Examples

1. 23.5 – Three significant digits as all are non-zero numbers (see rule #1 above).

2. 23.50 – Four significant digits. The final zero is significant because the number contains a decimal place (see rule #4 above).

3. 402 – Three significant digits. The zero is between non-zero numbers (see rule #2 above).

4. 5,200 – Two significant digits. There is no decimal place so the trailing zeros are simply placeholders and not-significant (see rule #4 above).

5. 0.030 – Two significant digits. Leading zeros are never significant (see rule #3 above). The trailing zero is significant because the number contains a decimal place (see rule #4 above).

6. 0.0070080 – Five significant digits. Leading zeros are never significant (see rule #3 above). The two zeros between 7 and 8 are significant because they are between non-zero numbers (see rule #2 above). The trailing zero is significant because the number contains a decimal place (see rule #4 above).

For additional practice problems on significant digits and measurements, visit Significant Digits & Measurements Practice Problems.

[Download the accompanying PDF worksheet.]

1. A bright orange, crystalline substance is analyzed and determined to have the following mass percentages: 17.5% Na, 39.7% Cr, and 42.8% O. Determine its empirical formula.

2. A sample of cisplatin is 65.02% platinum, 9.34% nitrogen, 2.02% hydrogen, and 23.63% chlorine. Determine the empirical formula.

3. What are the empirical formula and the molecular formula for resorcinol? It is 65.44% carbon, 5.49% hydrogen, and 29.06% oxygen with a molar mass of 110 grams/mole.

4. Caffeine contains 49.5% C, 5.15% H, 28.9% N and 16.5 % O by mass and the molar mass is about 195 g/mol. What are its empirical and molecular formulas?

[Download the accompanying PDF worksheet.]

Perform the following calculations, being sure to give the answer with the correct number of significant digits.

1. How many atoms are in 8.21 g Li?
2. What is the mass in grams of 4.17×10²¹ formula units of magnesium perchlorate?
3. How many sulfur atoms are in 72 g of cobalt(III) sulfate?
4. How many grams of lithium are in 30.6 g of lithium nitride?
5. A sample of cadmium borate contains 72.9 g of boron.  What mass of cadmium will the sample contain?
6. What is the mass of sulfur in grams in a sample of manganese(VI) sulfate containing 8.291×10²⁴ oxygen atoms?

[Download the accompanying PDF worksheet.]

Solve the following problems:

1. What is the mass in grams of 51.2 moles of trichlorine nonafluoride?
2. What is the mass in grams of 3.00 moles of magnesium chlorite?
3. How many moles are in 189 grams of calcium hypoarsenite?
4. How many moles are in 0.0395 grams of cobalt(VI) permanganate?

What is Molar Mass?

Molar mass is literally the mass of one mole of a substance. We saw in a previous lesson on atomic mass that the periodic table provides us with the average atomic mass of elements. For example, the atomic mass of hydrogen (symbol H) is 1.01 amu (atomic mass units). But one incredibly handy aspect of the periodic table is that those same masses are also the mass of one mole of the element in grams.

For example, one atom of hydrogen has a mass of 1.01 amu, but one mole of hydrogen has a mass of 1.01 g. Remember that one mole is equal to Avogardo’s number, 6.022×10²³ particles.  So one mole of hydrogen, or 6.022×10²³ hydrogen atoms, has a mass of 1.01 g.

Molar Mass of Compounds

To find the mass of one mole of any substance we count the elements present and add up their atomic masses.

For example, let’s calculate the molar mass of iron(III) oxide, Fe₂O₃. Fe₂O₃ is composed of two iron atoms and three oxygen atoms. To calculate its molar mass we have to add up the mass of two iron atoms and three oxygen atoms.

2 x Fe  =  2 x 55.85  =  111.70

3 x O   =  3 x 16.00  =  + 48.00

Total                       =   159.70 g/mol

This is the molar mass of Fe₂O₃. So one mole of Fe₂O₃ would have a mass of 159.70 g. Or Avogadro’s number, 6.022×10²³, of Fe₂O₃ formula units (ionic compound particles) would have a mass of 159.70 g.

[In terms of significant digits, we normally just keep as many decimal places for the total mass as was originally present in the mass of the element.  For example, iron (Fe) has a mass of 55.85 - two decimal places.  So when we add up two iron atoms, we keep two decimal places and write 111.70.]

Examples

Calculate the molar mass of:

1. Mn(MnO₄)₄

There are 5 Mn atoms (mass of 54.94) and 16 O atoms (mass of 16.00).

5 x 54.94  +  16 x 16.00  =  530.70 g/mol

2. Al₂(SO₄)₃

There are 2 Al atoms (mass of 26.98), 3 S atoms (mass of 32.07), and 12 O atoms (mass of 16.00).

2 x 26.98  +  3 x 32.07  +  12 x 16.00  =  342.17 g/mol

Notice that the units on molar mass are grams per mole (g/mol). This looks a lot like a conversion factor that relates two units: grams to moles. Just like there are 12 inches/1 foot, for Al₂(SO₄)₃ there are 342.17 g/mol. This allows us to convert between grams of Al₂(SO₄)₃ and moles of Al₂(SO₄)₃. The following two example problems will illustrate how to use molar mass as a conversion factor between moles and mass.

Example Problems

3. What is the mass in grams of 3.81 moles of Fe₂O₃?

We earlier calculated the molar mass of Fe₂O₃ to be 159.70 g/mol.

3.81 mol Fe₂O₃  x  159.70 g/1 mol Fe₂O₃  =  608 g

4.  How many moles are in 713 grams of Al₂(SO₄)₃?

The molar mass of Al₂(SO₄)₃ was earlier calculated to be 342.17 g/mol. Since we are converting grams into moles, we need to flip this conversion factor so that grams are on the bottom in order to cancel.

713 g  x  1 mol Al₂(SO₄)₃/342.17 g  =  2.08 mol Al₂(SO₄)₃

[Get more Practice Problems on Molar Mass Conversions here.]

[Download the accompanying PDF worksheet here.]

Name the following ionic compounds, molecular compounds, and acids:

1. Al₂S₃
2. H₃PO₃(aq)
3. P₂O₅
4. Cu(OH)₂
5. HClO₄(aq)
6. K₂O
7. (NH₄)₂CO₃
8. HC₂H₃O₂(aq)
9. Ni(SO₄)₂

[Download the accompanying PDF worksheet here.]

Determine the formulas of the following ionic compounds, molecular compounds, and acids:

1. lead(II) chloride
2. magnesium phosphate
3. nitrogen triiodide
4. hydrofluoric acid
5. calcium nitride
6. tin(IV) oxide
7. dichlorine heptaoxide
8. sodium hydrogen sulfate
9. triantimony hexasulfide

[Download the accompanying PDF worksheet here.]

Write formulas for the following ionic compounds:

1. barium sulfide
2. aluminum selenate
3. manganese(V) oxide
4. beryllium hypoarsenite
5. gallium phosphide
6. iron(II) phosphite
7. platinum(II) cyanide
8. mercury(II) hydrogen carbonate
9. lithium perbromate

What Are Polyatomic Ions?

Unlike monoatomic ions, which are composed of a single atom, polyatomic ions are composed of many atoms.  (The prefix poly- means “many,” as in polygon or polytheism.)

There are some rules and patterns which can help us to deduce the structures of these ions, but before we can learn those, there are a handful that don’t follow the patterns that you just need to memorize.  Depending on your course or instructor, you may need to memorize some of the following.  When it comes to memorizing the, flashcards are your friends!

The good news, is there are dozens of other polyatomic ions that are relatively easy to figure just by knowing or observing a few patterns.

The -ate Ions

Many of the nonmetals combine with oxygen to make anions (negatively charged ions) that we are sometimes referred to as oxyanions. If you refer to the periodic table below, you will that a several nonmetals have been highlighted that commonly form such oxyanions.

These nonmetals for polyatomic ions that end in the suffix -ate. However, the number of oxygens and the charge on the ions varies as we can see in the following table.

Patterns for Polyatomic Oxyanions

1. Notice how the four polyatomic-ate ions in the center square (phosphate, arsenate, sulfate, and selenate) all have four oxygen atoms, while the polyatomic -ate ions on the outside all have three oxygen atoms.
2. As you start from right-hand side, the first column of polyatomic -ate ions (chlorate, bromate, iodate) all have a 1- charge. The second column (sulfate, selenate) all have a 2- charge, while the third column (phosphate, arsenate) all have a 3- charge. On the top row, the farthest right polyatomic -ate ion is nitrate which is 1-, followed by carbonate which is 2-, followed by borate which is 3-.
3. Within a column, the pattern also repeats.  For example, if you know that chlorate is ClO₃^-, then you can predict that all of the other oxyanions made from nonmetals in group 17/VIIA follow the same pattern. Cl combines with three oxygens and has a 1- charge; Br combines with three oxygens and has a 1- charge; I combines with three oxygens and has a 1- charge. In group 16/VIA, both S and Se have the same pattern. In group 15/VA, P and As have the same pattern, but notice N is different, combining with only three oxygens and having only a 1- charge.

Changing the Number of Oxygens

Once you know the polyatomic ions that end in the suffix -ate there are only a few more patterns to know and you’ll be on your way to naming dozens and dozens of polyatomic ions!

For example, let’s use chlorate as an example. We know that chlorate is ClO₃^-.  This is the anion that ends in -ate. Let’s look at the chart above and see if we can figure out what perchlorate would be.  Since it has the prefix per- and the suffix -ate, it must have one O atom more than chlorate, meaning the formula is ClO₄^-. Notice only the number of oxygen atoms changed, the charge did not.

What would chlorite be?  The suffix -ite means it has one O atom less than chlorate, so it must be ClO₂^-.

How about hypochlorite?  The prefix hypo- and the suffix -ite mean it has two less O atoms than chlorate, so it must be ClO^-.

We can similarly predict sulfite and hyposulfite once we know the formula and charge of sulfate. (Ignore persulfate as it’s an exception to the general rule.)

Notice that the prefixes and suffixes don’t denote the total number of O atoms, just that it has more or less than the -ate polyatomic ion.

Example Problems

Predict the formula and charge for:

1. sulfite ion – We determine by sulfite is by first figuring out what sulfate.  Sulfate comes from sulfur which is in the center box of the nonmetals so it have four oxygens. It’s two columns away from the right-hand side so its charge is 2-. Sulfate is SO₄^2-. The suffix -ite tells me it lost one oxygen atom, so sulfite must be SO₃^2-.

2. hypophosphite ion – Let’s figure out what phosphate is. The root comes from phosphrous which is in the center box of nonmetals so it must have four oxygen atoms. Phosphrous is three columns away from the right-hand side so its charge is 3-. Therefore, phosphate is PO₄^3-. The prefix hypo- and the suffix -ite tell me to subtract two oxygen atoms, so hypophosphite must be PO₂^3-.

3. perbromate ion – Bromate comes from bromine which is outside the center box of nonmetals, so it combines with three oxygens.  It’s in the first column from the right so its charge is 1-. Bromate must be BrO₃^-.  The prefix per- tells me to add an oxygen, so perbromate must be BrO₄^-.

4. nitrite ion – Nitrate is NO₃^- and the -ite suffix means I must remove an oxygen atom, so nitrite must be NO₂^-.

How would you name the following:

5. IO₂^- – Let’s determine what the -ate version of iodine’s polyatomic is. Iodine is outside the center box, so it must have three oxygen atoms.  Iodine is in the first column from the right-hand side so the charge on the polyatomic must be 1- (or just -). Therefore, iodate must be IO₃^-. However, this polyatomic has one less oxygen atom which means the suffix changes to -ite, so we would name it iodite ion.

6. AsO₂^3- – Arsenate would be AsO₄^3- and this has two fewer oxygen atoms. According to our rules, two less oxygens means we add the prefix hypo- and change the suffix to -ite, so this would be the hypoarsenite ion.

7. SeO₃^2- – Selenate is SeO₄^2-. The given ion has one less oxygen atom, so we must change the suffix from -ate to -ite. Therefore, we would name this the selenite ion.

“Hydrogen” Polyatomic Ions

Another way to form polyatomic ions is by combining them with one or more hydrogen ions, H⁺. For example, we could combine H⁺ with carbonate, CO₃^2- to form hydrogen carbonate, HCO₃^-. Notice the overall charge is 1- because the 1+ on H⁺ combine with the 2- on CO₃^2-.

What would hydrogen sulfate be? “Hydrogen” tells me to add H⁺ to sulfate, SO₄^2-. So hydrogen sulfate would be HSO₄^-.

When we add hydrogen ions to a polyatomic with a 3- charge, such as phosphate (PO₄^3-), we could add either one or two H⁺ ions and it would still remain a polyatomic ion. (Adding three H⁺ would completely cancel out the 3- charge making it a neutral compound and no longer a polyatomic ion.)  We have two options then, HPO₄^2- and H₂PO₄^-. The former has only one hydrogen ion so call it either hydrogen phosphate ion, or sometimes monohydrogen phosphate ion.  The latter has two hydrogen ions so we refer to it as dihydrogen phosphate ion.

Example Problems

Predict the formula and charge for:

8. hydrogen hyposulfite – Sulfate is SO₄^2- so hyposulfite tells me to reduce the number of oxyen atoms by two, leaving SO₂^2-. Adding a hydrogen ion, H⁺, it becomes HSO₂^-.

9. hydrogen selenide - The suffix -ide on selenide tells me the ion probably is a monoatomic anion formed from a nonmetal, in this case from selenium. Selenium is two columns away from the noble gases so it needs two electrons, meaning selenide is Se^2-. If I add H⁺ to this, I get HS^-.

10. dihydrogen borate – Borate is BO₃^3-. Dihydrogen tells me to add two H⁺ ions, which produces H₂BO3^-.

How would you name:

11. HSO₄^- – If I remove an H⁺ from the ion, I am left with SO₄^2-. (The charge becomes 2- because I removed a positive hydrogen ion.)  SO₄^2- is named sulfate so this ion combined with hydrogen would be hydrogen sulfate ion.

12. H₂AsO₃^- – Removing two H⁺ ions leaves me with AsO₃^3-. I know that arsenate is AsO₄^3- and this has one less oxygen atom, so it must be aresnite. I’ve added two hydrogens to the polyatomic ion, so the result is dihydrogen arsenite ion.

[Download the accompanying PDF worksheet.]

1. Magnesium has three naturally occurring isotopes, ²⁴Mg (78.99%, 23.9850 amu), ²⁵Mg (10.00%, 24.9858 amu), and ²⁶Mg.  Determine the percent abundance and isotopic mass of ²⁶Mg.

2. Boron has two naturally occurring isotopes.  Find the percent abundances of ¹⁰B and ¹¹B given the isotopic mass of ¹⁰B = 10.0129 amu and the isotopic mass of ¹¹B = 11.0093 amu.

3. Chlorine has two naturally occurring isotopes, ³⁵Cl (34.9689 amu) and ³⁷Cl (36.9659 amu).  If chlorine has an average atomic mass of 35.4527 amu, what is the percent abundance of each isotope?