The post Chemistry Lesson: Average Atomic Mass Calculations appeared first on Get Chemistry Help.
]]>The atomic mass is an experimental number determined from all of the naturally occuring isotopes of an element. As we saw in our lesson on atomic structure, not all atoms of an element are identical. For example, hydrogen has three different isotopes that occur in nature - ^{1}H, ^{2}H, ^{3}H. So when you look on the periodic table and see that it has an atomic mass of 1.01 amu, that is the average of the masses of all three isotopes, not just one of them.
Notice that the units were listed as amu, which stands for atomic mass units. This unit is based off the mass of the isotope ^{12}C (carbon-12). Carbon-12 was chosen as the basis for all of the masses on the periodic table and has been defined to be exactly 12 amu. So all of the other masses on the period table are relative to the mass of carbon-12. In that case, then why is the atomic mass of carbon on the periodic table not exactly 12 amu? Because not all carbon in nature is ^{12}C! Most of it is ^{12}C, some of it is ^{13}C, and a very tiny amount is ^{14}C. When these are averaged together you get the average atomic mass shown on the periodic table of 12.01 amu.
Since the abundances are not equal, we cannot do a typical simple average where we just add them up and divide by three. Instead, we need to perform a weighted average. The formula to calculate the average atomic mass is:
average atomic mass = ∑(relative abundance x mass of isotope)
Remember that ∑ is the symbol for sum. In other words, we will take the sum of the relative abundance of each isotope multipled by its mass.
Neon has three naturally occuring isotopes.
Symbol | Mass number | Isotopic mass (amu) | Percent natural abundance |
Ne-20 | 20 | 19.9924 | 90.48% |
Ne-21 | 21 | 20.9938 | 0.27% |
Ne-22 | 22 | 21.9914 | 9.25% |
Remember that mass number is not the same as the atomic mass or isotopic mass! The mass number is the number of protons + neutrons, while atomic mass (or isotopic mass) is the mass if you were to somehow weigh it on a balance.
To find the average atomic mass of neon, we will use the equation above and take the abundance of the first isotope times the mass of the first isotope plus the abundance of the second isotope times the mass of the second isotope plus the abundance of the third isotope times the mass of the third isotope. However, you might recall from your math courses that when you use a percentage in a calculation you always want to use the decimal form, meaning you must first divide the percentage by 100. The equation would then look like:
= (0.9048 x 19.9924 amu) + (0.0027 x 20.9938 amu) + (0.0925 x 21.9914 amu)
= 18.089 amu + 0.0516 amu + 2.034 amu
= 20.1796 amu
Looking at our significant digits, we see that 18.089 was precise to the hundredths place, 0.0516 was precise to the thousandths place, and 2.034 was precise to the hundredths place. So the answer can only be precise out to the least precise or most uncertain place, hundredths place. We would therefore round the answer off to 20.18 amu. This is the average atomic mass of neon.
Let’s just take a second and see if 20.18 amu seems like a reasonable answer, given the initial data. According to the table, most neon (over 90%) has a mass of approximately 20 amu, a tiny bit has a mass of around 21 amu, and around 9% has a mass of 22 amu. Based off this we would predict that the answer would be close to 20 amu, but slightly higher, which it is!
If silver is 51.84% Ag-107 with a mass of 106.9051 amu and the rest Ag-109 with a mass of 108.9048 amu, calculate silver’s atomic mass.
Just like before, we need to take the abundance of Ag-107 times the mass of Ag-107 plus the abundance of Ag-109 times the mass of Ag-109. But notice that while we have the abundance of Ag-107, 51.84%, we are not given the abundance of Ag-109. How can we figure it out? Well, both abundances have to total 100% so is Ag-107 is 51.84%, then Ag-109 must be 100 – 51.84 = 48.16%.
= (0.5184 x 106.9051 amu) + (0.4816 x 108.9048 amu)
= 55.240 amu + 52.449 amu
= 107.869 amu
Looking at the significant digits, both 55.240 amu and 52.449 amu are precise to the hundredths place so our answer must also be rounded off to the hundredths place. The average atomic mass of silver is 107.87 amu.
[For additional practice problems on calculating atomic mass, visit Atomic Mass Practice Problems Part I.]
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]]>The post Chemistry Lesson: The Metric System & Conversions appeared first on Get Chemistry Help.
]]>The metric system was created by the French in the late 1700′s in response to the English system which, quite frankly, is fairly confusing in that it uses numerous different units and has no uniformity among conversions. For example, in the English system 1 ft = 12 in., 1 yd = 3 ft, and 1 mile = 5280 ft. But over the course of about 10 years a French committee created a standardized uniform system known as the metric system.
Unlike the English system which has multiple units for length (inch, foot, yard, mile), the metric only has one – the meter (m). Similarly, there are four basic metric units:
Length – meter (m)
Mass – gram (g)
Volume – liter (L)
Time – second (s)
The metric system is a decimal system. Prefixes are used to increase and decrease one of the basic units by a factor of 10 as seen on the following metric table.
Prefix | Symbol | Magnitude | Meaning (multiply by) |
tera- | T | 10^{12} | 1 000 000 000 000 |
giga- | G | 10^{9} | 1 000 000 000 |
mega- | M | 10^{6} | 1 000 000 |
kilo- | k | 10^{3} | 1 000 |
– | – | – | – |
deci- | d | 10^{-1} | 0.1 |
centi- | c | 10^{-2} | 0.01 |
milli- | m | 10^{-3} | 0.001 |
micro- | μ | 10^{-6} | 0.000 001 |
nano- | n | 10^{-9} | 0.000 000 001 |
pico- | p | 10^{-12} | 0.000 000 000 001 |
femto- | f | 10^{-15} | 0.000 000 000 000 001 |
On the metric table above, the line represented by — is the base unit. Prefixes above this line increase the size of the base unit, while prefixes below this line decrease the size of the base unit.
1. kilogram – Kilo (k) is the prefix and gram (g) is the base unit, so the abbreviation is kg. But what is a kilogram? According to the table above, kilo means 10^{3} so 1 kg = 10^{3} g.
Important: When converting metric units in to base units, the prefix and the exponent will always be on different sides of the conversion factor. In the above example, this means that since kilo (k) is on one side, 10^{3} must be on the opposite side. Using the table above, the prefix and the exponent will never be on the same side of a metric conversion.
2. gigameter - Giga (G) is the prefix and meter (m) is the base unit, so gigameter is abbreviated Gm. (Notice that giga is a capital G, while gram is a lower-case g.) What is a gigameter? 1 Gm = 10^{9} m
3. microliter - Micro (μ) is the prefix and liter (L) is the base unit, so a microliter is written as μL. 1 μL = 10^{6} L
4. millisecond - Milli (m) is the prefix and second (s) is the base unit, so the abbreviation is ms. 1 ms = 10^{-3} s
Convert 4.29 ng into grams.
We are being asked to convert ng to grams so we need a metric conversion that relates the two. What does ng mean? According to the chart above, nano (n) =10^{-9}, so 1 ng = 10^{-9} g. When using this as a conversion factor we want the initial given units, ng, to cancel so we need to ensure that it is on the bottom.
4.29 ng x 10^{-9} g/1 ng = 4.29×10^{-9} g
Remember that metric-metric conversions have been defined and are exact, therefore they do not affect the significant digits of the final answer. Since the original measurement, 4.29 ng had three significant digits, the final answer will as well.
Convert 7.624×10^{3} cm into Mm.
There is no direct metric conversion between centi- (c) and mega- (M), but both share the same base unit of meters (m). Because of this, we can use meters as the crossroads and convert cm -> m -> Mm.
To begin with, we need to convert cm into m. Referring to the above table, 1 cm = 10^{-2} m. Our starting unit is cm so we need cm in the denominator of the conversion factor in order for it to cancel.
7.624×10^{3} cm x 10^{-2} m/1 cm
Centimeters has canceled and we are left with meters (m) which we need to convert to Mm. 1 Mm = 10^{6} m. Since we are in meters, we need to ensure that 10^{6} m is on the bottom of the conversion factor so that it will cancel.
7.624×10^{3} cm x 10^{-2} m/1 cm x 1 Mm/10^{-6} m = 7.624×10^{-5} Mm
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]]>The post Chemistry Lesson: Unit Analysis (Unit Conversions, Dimensional Analysis) appeared first on Get Chemistry Help.
]]>When we need to convert between units (for example, converting minutes to seconds) we use conversion factors. Conversion factors are a ratio that specify how one unit of measurement is related to another unit of measurement. For example, when converting minutes to seconds we can use the relationship 1 minute = 60 seconds. This can be written as two ratios or two different conversion factors:
1 minute/60 seconds OR 60 seconds/1 minute
How many seconds are in 3.55 minutes?
1. Identify the known quantity and the units of the new quantity.
Known: 3.55 minutes
Units of New Quantity: seconds
2. Multiply the given quantity by one or more conversion factors so that units cancel, leaving only the desired units.
3.55 minutes x = seconds
Which conversion factor will cause minutes to cancel and leave us with seconds: 1 minute/60 seconds OR 60 seconds/1 minute? In order for minutes to cancel we need minutes in the denominator so it must be 60 seconds/1 minute.
3.55 minutes x 60 seconds/1 minute =
3. Perform the mathematical operations indicated by the conversion factor setup.
3.55 minutes x 60 seconds/1 minute = 213 seconds
English-English and metric-metric conversions are always exact. In other words, when converting from one unit to another unit in the same system (feet to inches, meters to centimeters), the conversions have been defined and are not measurements so they do not affect the significant digits of the calculation.
1 ft = 12 in. 1 yd = 3 ft 1 gal =4 qt
1 km = 10^{3} m 1 cg = 10^{-2} g 1 GL = 10^{9} L
However, English-metric conversions are inexact. They have been measured and not defined so the significant digits do affect the significant digits of the calculation.
1 lb ≈ 454 g (3 sig figs) 1 qt ≈ 0.9463 L (4 sig figs)
The only exact English-metric conversion is 1 in. = 2.54 cm which was defined in 1958. This conversion will not affect the significant digits of the calculation.
How many yards are in 2.7 miles? (1 yd = 3 ft, 1 mile = 5280 ft)
We are beginning with 2.7 miles so we need a conversion factor with miles in it and miles will need to be on the bottom in order for it to cancel. So we begin setting up the calculation like this:
2.7 miles x 5280 ft/1 mile
Now we have feet but want yards, so we need a conversion factor that will put ft in the denominator (bottom) and yards in the top (numerator):
2.7 miles x 5280 ft/1 mile x 1 yd/3 ft = 4752 yd
Looking at the significant digits, the original measurement of 2.7 miles had 2 significant digits. In the first conversion factor, is 5280 ft considered 3 significant digits or exact? Since the conversion is between two English units it is exact, meaning it will not affect the final significant digts of the answer. 1 yd = 3 ft is also English-English meaning this factor is exact. Thus, the only significant digits to consider are the 2 signifnicant digits in the original measurement meaning the answer should be rounded off to 2 significant digits: 4800 yd or the less ambiguous way, 4.8 x 10^{3} yd
How many kL are in 3.190 x 10^{5} gal? (1 L ≈ 0.264 gal, 1 kL = 10^{3} L)
In order for the original gallons to be converted in to kL, we will need a setup that looks like:
3.190 x 10^{5} gal x 1 L/0.264 gal x 1 kL/10^{3} L = 1208.33 kL
How many significant digits should the final answer have?
The original measurement (3.190 x 10^{5} gal) has 4 significant digits. In the first conversion factor, we are converting 0.264 gal (English) to 1 L (metric) so it must be inexact, meaning 0.264 is considered 3 significant digits. In the second conversion factor, we are converting L (metric) to kL (metric) so this is an exact conversion, meaning it will not affect the significant digits of the final answer in the calculation. Therefore, the answer must have the fewest or 3 significant digits: 1.21 x 10^{3} kL
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]]>The post Chemistry Lesson: Structure of the Atom – Protons, Neutrons & Electrons appeared first on Get Chemistry Help.
]]>The atom is very tiny, having a diameter of about 10^{-10} meters. Keep in mind that’s like saying 0.0000000001 meters, or about one-billionth of a meter across! But within this already extremely small atom is an even smaller region called the nucleus. The nucleus has a diameter of 10^{-14} m, so it is about 10,000 times smaller than the volume of the atom as a whole. One analogy is to imagine the atom is the size of a football stadium. In that gigantic stadium the nucleus would only be the size of a small marble sitting on the 50-yard line!
If this were the volume of the atom… | then this would be the volume of the nucleus. |
Within the atom are three basic subatomic particles: protons (p^{+}), neutrons (n^{0}), and electrons (e^{-}). We abbreviate the proton as p^{+} because it has a positive charge, the neutron as n^{0} because it is neutral, and the electron as e^{-} because it has a negative charge.
Mass (kg) | Mass (amu) | Charge | |
proton | 1.67262×10^{-27} | 1.0073 | 1+ |
neutron | 1.67493×10^{-27} | 1.0087 | 0 |
electron | 0.00091×10^{-27} | 0.00055 | 1- |
The protons and neutrons live in the nucleus, while the electrons are orbiting the nucleus similar to the way planets orbit a star. (We’ll see in a later video the various orbits these electrons can occupy.) In a neutral atom, the number of positively-charge protons must equal the number of negatively-charged electrons. If there were more protons than electrons, then it would have a positive charge overall. While if it had more electrons than protons, it would have an overall negative charge. Atoms with positive or negative charges are called ions. But for atoms they are always neutral and the number of protons equals the number of electrons.
In terms of mass, we see that protons and neutrons are very similar with both having a mass of approximately 1 amu (atomic mass units). Now compare that to the mass of the electrons which are about 1,800 times less massive. As an analogy, if protons and neutrons had a mass around that of a baseball, the mass of an electron would about that of a single grain of rice!
If protons and neutrons had the mass of baseballs… | then electons would have the mass of a grain of rice. |
But where do the massive protons and electrons live? In the tiny nucleus. This means that rather than the mass being spread out evenly across the atom, 99.9% of the mass of an atom is contained in just 1/10,000th of its volume. Going back to the football stadium analogy, it’s as if 99.9% of the mass of that huge stadium was condensed down in to the tiny marble sitting on the 50-yard line. Because so much mass is contained in such a tiny volume this makes the nucleus incredibly dense! For comparison, let’s look at a few other substances that are commonly thought of as dense.
Density (g/cm^{3}) | |
Lead | 11.34 |
Gold | 19.30 |
Osmium | 22.59 |
Nucleus | 10^{13}-10^{14} |
Lead is very dense which is why it is often used to block radiation. Its density is 11.34 g/cm^{3}. This means that if you had a cube that was 1 cm x 1 cm x 1cm it would have a mass of 11.34 g. By comparison, the same sized cube of gold would weigh 19.30 g. So a cube of gold would be much heavier than the same sized cube of lead. The most dense element we know of is osmium whose density is 22.59 g/cm^{3}. This means that a 1 cm x 1 cm x 1 cm cube of osmium would be heavier than gold and about twice as heavy as a cube of lead.
But all of these densities pale in comparison to the incredibly dense nucleus. By compressing 99.9% of the mass of the atom in to just 1/10,000th of the volume the density of a typical nucleus is between 10-100,000,000,000,000 g/cm^{3}! So the same 1 cm x 1 cm x 1 cm cube would have a mass between ten and one hundred trillion grams! We cannot fathom how dense this truly is. In fact, if you were able to collect enough nuclei to fill a small matchbox it would have a mass of over 2.5 billion tons! Think about that for a second: this tiny matchbox would have a mass equal to that of a billion cars!^{ }
= one billion cars |
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]]>[Download the accompanying PDF worksheet.]
Convert the following numbers into scientific notation:
1. 192 Gs
2. 0.00478 fm
3. 50800 qt
4. 390.0200 K
5. 0.08200 µm
6. 4000 °F
Convert the following numbers into standard (decimal) notation:
7. 5.80 × 10^{-2} ft
8. 9.0 × 10^{-4} pL
9. 7.495 × 10^{6} gal
10. 4.1070 × 10^{2} hr
11. 1.02 × 10^{-6} Ts
12. 8.28 × 10^{1} in.
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]]>Scientific notation is a method of writing really large or really small numbers in a more concise form that removes all of the extraneous zeroes. For example, 0.0000000072 becomes 7.2 × 10^{-9} and 14,000,000 becomes 1.4 × 10^{7}.
But first, we need to remind ourselves how exponents work. Exponents indicate that a number has been multiplied by itself. For example, we could write 7 × 7 × 7 as 7^{3}. The superscript 3 indicates that 7 was multiplied by itself three times. We call this “seven to the third power” or “seven cubed”. If you divide by the same number multiple times, like 1/4 × 1/4 × 1/4, then the exponent is 4^{-3}. The negative sign shows that you divided by 4 three times.
In scientific notation, we use the basic format D.DD × 10^{n}. D.DD is called the coefficient and represents the significant digits in a number. This is multiplied by the base, which is scientific notation is always 10. So when converting a number in “standard” or “decimal” notation to scientific, there are two basic steps.
Step 1: Put the decimal after the first significant digit.
Step 2: Indicate how many places the decimal moved by the power of 10.
- A positive power of 10 indicates the decimal moved to the left.
For example, 300,000,000 m/s becomes 3 × 10^{8} m/s
- A negative power of 10 indicates the decimal moved to the right.
For example, 0.000000140 m becomes 1.40 × 10^{-7} m
Scientific Notation Examples
Convert the following numbers from standard notation to scientific notation:
1. The distance from the earth to the moon is 238,900 miles.
First we pull out the significant digits from 238,900. The trailing zeroes are not significant because there is no decimal place so the significant digits are 2389. Next, we want to move the decimal until it is located just after (to the right of) the first significant digit, so 2.389. How many times did we have to move the decimal to get it from it’s understood or assumed location after the last zero in 238,900 to just after the 2? It moved five times to the left so we write the scientific notation as 2.389 × 10^{5} miles. This is correct because if we actually multiplied 2.389 by ten five times we would in fact get the original number of 238,900.
2. The mass of a dust particle is 0.000 000 000 753 kg.
The zeroes to the left of 753 are not significant because leading zeroes never are. Thus, only 753 is significant. We place the decimal after the seven giving us 7.53. How many times does to the decimal how to move from its original position to get just after the 7? It has to move ten places to the right, so the answer is 7.53 × 10^{-10} kg. The 10^{-10} basically means that you are now dividing 7.53 by ten, ten times which would in fact give us the original number.
Convert the following numbers from scientific notation to standard notation:
3. The hydrogen-oxygen bond in water is 9.584 × 10^{-11} m.
When we go from scientific notation back to standard we are doing the reverse of above. The × 10^{-11} tells us we are dividing by ten eleven times. This means the number must be very small so the decimal must move to the left eleven times. The answer is 0.000 000 000 095 84 m.
4. Mount Everest has a height of 2.9029 × 10^{4} ft.
We take 2.9029 and multiply it by ten four times. This makes the number larger so we move the decimal to the right four places. The answer is 29,029 ft.
5. A carbon atom weighs 2.00 × 10^{-23} grams.
The exponent tells us we are dividing by ten twenty-three times, so this must be a very small number! Therefore, let’s move the decimal twenty-three places to the left. The final, very small answer, is 0.000 000 000 000 000 000 000 020 0 grams. Notice the two zeroes on the end because the original coefficient was 2.00 and the decimal places told us that those ending zeroes are indeed significant.
For additional practice problems on scientific notation, visit Scientific Notation Practice Problems.
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]]>The post Chemistry Lesson: Significant Digits & Measurements appeared first on Get Chemistry Help.
]]>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.
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.
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 2.5×10^{2} 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.
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.
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]]>[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?
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]]>The post Chemistry Practice Problems: Mole Calculations appeared first on Get Chemistry Help.
]]>[Download the accompanying PDF worksheet.]
Perform the following calculations, being sure to give the answer with the correct number of significant digits.
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]]>The post Chemistry Practice Problems: Molar Mass Conversions appeared first on Get Chemistry Help.
]]>[Download the accompanying PDF worksheet.]
Solve the following problems:
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]]>