Calculate the isotopic abundances when given the average atomic weight and the isotopic weights

Let's start by repeating the solution for nitrogen from the Average Atomic Weight tutorial:

Lead, isotope of mass 210 Pb CID 6328175 - structure, chemical names, physical and chemical properties, classification, patents, literature, biological activities. Lead is a chemical element with the symbol Pb (from the Latin plumbum) and atomic number 82. It is a heavy metal that is denser than most common materials. Lead is soft and malleable, and also has a relatively low melting point. Chemical elements listed by atomic mass The elemenents of the periodic table sorted by atomic mass. Click on any element's name for further information on chemical properties, environmental data or health effects. This list contains the 118 elements of chemistry.

(14.003074) (0.9963) + (15.000108) (0.0037) = 14.007

The solution is laid out like this:

(exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element

In the other tutorial, the average atomic weight is the unknown value calculated. In this tutorial, the unknown values calculated are the TWO percent abundances.

However, you might protest that we have two unknowns, but only one equation. You would be right, except that there is a trick. Ah, the ChemTeam loves tricks!!

Think about the sum of the percent abundances of the TWO isotopes. That's right, they add up to 100% (or, since we use decimal abundances in the calculation, 1.00). So, the trick is to express both abundances using only one unknown. I'll show how using the above nitrogen example.

Problem #1: Nitrogen is made up of two isotopes, N-14 and N-15. Given nitrogen's atomic weight of 14.007, what is the percent abundance of each isotope?

Here's the solution:

What Element Is 205 83 Pb

(14.003074) (x) + (15.000108) (1 − x) = 14.007

Notice that the abundance of N-14 is assigned 'x' and the N-15 is 'one minus x.' This is the 'trick' refered to above. The two abundances always add up to one (or, if you prefer, 100%)

Comment One (see end of problem for another comment): sometimes the exact weights of the isotopes are not provided in the problem. However, in this Internet era, it is easy to look up the values on-line. I advise you to ask your instructor what the question wants you to do if only the mass numbers are provided in the problem. Please notice that the mass numbers (14 and 15) are quite close to the exact values. The consequence of using 14 and 15 rather than the exact values is that we will get slightly approximate answers for the two abundances.

For example, I might have this:

(14) (x) + (15) (1 − x) = 14.007

Solving gives:

14x + 15 − 15x = 14.007

After some more simple algebra, we get:

x = 15 − 14.007 = 0.993

so

1 − x = 0.007

You may want to compare them to the more exact abundances which are in the equation set up at the top of the page.

Comment Two: only the mass numbers may be given if the element is a made-up one. For example: 'ChemTeam-42 and ChemTeamium-44 exist. Calculate the abundances if the atomic weight is 42.68.' Since ChemTeamium doesn't exist, you can't look up exact isotopic weights. So, use 42 and 44 in the solution to the problem.

Problem #2a: Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given copper's atomic weight of 63.546, what is the percent abundance of each isotope?

Solution:

1) Write the following equation:

(62.9296) (x) + (64.9278) (1 − x) = 63.546

Once again, notice that 'x' and 'one minus x' add up to one.

2) Solve for x:

x = 0.6915 (the decimal abundance for Cu-63)

Note that this calculation technique works only with two isotopes. If you have three or more, there are too many variables and not enough equations. I hope it's obvious why you wouldn't do this with an element that has only one stable isotope!

By the way, the 'trick' works because the other equation required is:

x + y = 1

We simply went right to y = 1 − x and substitued it immediately without ever writing down the second equation required.

Problem #2b: Chlorine is made up of two isotopes, Cl-35 (34.969 amu) and Cl-37 (36.966 amu). Given chlorine's atomic weight of 35.453 amu, what is the percent abundance of each isotope?

1) Write the following equation:

(34.969) (x) + (36.966) (1 − x) = 35.453

2) Solve for x:

x = 0.7576 (the decimal abundance for Cl-35)

Problem #3a: A sample of naturally occurring silicon consists Si-28 (amu = 27.9769), Si-29 (amu = 28.9765) and Si-30 (amu = 29.9738). If the atomic mass of silicon is 28.0855 and the natural abundance of Si-29 is 4.67%, what are the natural abundances of Si-28 and Si-30?

Solution:

1) Set up a system of two equations in two unknowns:

Let x = isotopic abundance of Si-28 (as a decimal)
Let y = isotopic abundance of Si-30 (as a decimal)

Therefore:

(27.9769) (x) + (28.9765) (0.0467) + (29.9738) (y) = 28.0855

and

x + 0.0467 + y = 1.000

2) Rearrange the second equation to:

y = 1.000 − 0.0467 − x

then simplify it, substitute into the first equation and solve.

Note: isotopic abundances of the elements are well-known values. You can find the answers to Problem #3 here.

Problem #3b: Naturally occurring silicon consists of 3 isotopes, Si-28, Si-29 and Si-30, whose atomic masses are 27.9769, 28.9765 and 29.9738 respectively. The most abundant isotope is Si-28 which accounts for 92.23% of naturally occurring silicon. Given that the observed atomic mass of silicon is 28.0855 calculate the percentages of Si-29 and Si-30 in nature.

Solution:

1) Set up a system of two equations in two unknowns:

Let x = isotopic abundance of Si-29 (as a decimal)
Let y = isotopic abundance of Si-30 (as a decimal)

Therefore:

(27.9769) (0.9223) + (28.9765) (x) + (29.9738) (y) = 28.0855

and

0.9223 + x + y = 1.000

2) Rearrange the second equation to:

y = 1.000 − 0.9223 − x = 0.0777 − x

3) Substitute into the first equation and solve:

(27.9769) (0.9223) + (28.9765) (x) + (29.9738) (0.0777 − x) = 28.0855

Algebra!

x = 0.04668518 = 4.67%

Comment: when I first encountered this problem (Sept 2011), the given masses for Si-29 and Si-30 were 28.9865 and 29.9838. Notice the value in the 0.01 place and compare it to the ones used in the problem. It is my opinion that these small changes were made by the author of the problem in order to discourage someone from going to a source like Wikipedia and simply copying down the abundances given there.

Therein lies a warning to all students in this, the age of the Internet.

Problem #4: Determine the percent abundance for Fe-57 and Fe-58, given the following data:

Isotope	Atomic Weight	Percent Abundance
Fe-54	53.9396	5.845
Fe-56	55.9349	91.754
Fe-57	56.9354	???
Fe-58	57.9333	???

Solution:

1) Assign the percent abundance of Fe-57 to the variable 'x'

2) We need to get the percent abundance for Fe-58 in terms of x. Like this:

1 minus (0.05845 + 0.91754 + x)

the term (0.05845 + 0.91754 + x) is the total of the other three isotopic abundances.

The '1' represents 100% but expressed as a decimal value.

I will use (0.02401 minus x) just below.

3) Set up (and solve) the following equation:

(53.9396) (0.05845) + (55.9349) (0.91754) + (56.9354) (x) + (57.9333) (0.02401 − x) = 55.845

Algebra ensues.

x = 0.0213

The Fe-58 percent abundance is left to the reader. You can verify that 0.0213 is the correct answer by searching for 'isotopes of iron' on-line.

Note that the problem does not supply the average atomic weight of iron. This value is easily available, both in books and on the Internet.

Problem #5: Naturally occurring strontium consists of four isotopes, Sr-84, Sr-86, Sr-87 and Sr-88, whose atomic masses are 83.9134, 85.9094, 86.9089 and 87.9056 amu respectively. The most abundant isotope is Sr-88 which accounts for 82.58% of naturally occurring strontium and the least abundant isotope is Sr-84 which accounts for 0.56% of naturally occurring strontium. Given that the observed atomic mass of strontium is 87.62 amu, calculate the percentages of Sr-86 and Sr-87 in nature

Solution:

1) Assign the percent abundance of Sr-86 to the variable 'x'

2) We need to get the percent abundance for Sr-87 in terms of x. Like this:

1 minus (0.0056 + 0.8258 + x)

the term (0.0056 + 0.8258 + x) is the total of the other three isotopic abundances.

The '1' represents 100% but expressed as a decimal value.

I will use (0.1686 minus x) just below.

3) Set up (and solve) the following equation:

(83.9134) (0.0056) + (85.9094) (x) + (86.9089) (0.1686 − x) + (87.9056) (0.8258) = 87.62

Algebra ensues.

x = 0.09525

When you search for information about the isotopes of strontium, you will find that the abundance for Sr-86 is given as 0.0986. I may have made a mistake somewhere, probably in the calculation to get the value for x. I did not go back and double-check my calculations.

Problem #6: Oxygen is composed of three isotopes: One has a a mass of 16.999 amu. This isotope makes up 0.037% of oxygen. Of the other two, one has a mass of 15.995 amu, and the other has a mass of 17.999 amu. Calculate the abundance of the other two isotopes, using the average atomic mass of 15.9994 amu.

Solution:

1) Set abundances (as decimal percents):

O-16: x
O-17: 0.00037
O-18: 0.99963 − x

Note that 0.99963 is (1 minus 0.00037)

2) Set up average atomic weight equation:

(15.995) (x) + (16.999) (0.00037) + (17.999) (0.99963 − x) = 15.9994

Algebra ensues.

x = 0.9976

3) The percent abundances are as follows:

O-16: 99.76%
O-17: 0.037%
O-18: 0.203%

Problem #7: Pb has an average atomic mass of 207.19 amu. The three major isotopes of Pb are Pb-206 (205.98 amu); Pb-207 (206.98 amu); and Pb-208 (207.98 amu). If the isotopes of Pb-207 and Pb-208 are present in equal amounts, calculate the percent abundance of Pb-206, Pb-207, Pb-208.

Solution:

1) Set the percent abundances:

Pb-206: 1 − 2x
Pb-207: x
Pb-208: x

2) Set up average atomic weight equation:

(205.98) (1 − 2x) + (206.98) (x) + (207.98) (x) = 207.19

Algebra ensues.

x = 0.403

3) The percent abundances are as follows:

Pb-206: 19.4%
Pb-207: 40.3%
Pb-208: 40.3%

Here is another version of the above question: The atomic weight of lead is 207.19 amu. Assuming that it's made up of entirely the isotopes of Pb-206 (205.98 amu); Pb-207 (206.98 amu); and Pb-208 (207.98 amu) and that the percentages of the two lightest isotopes are equal, calculate the relative percentages of these isotopes in the natural element.

Solution:

1) Set up average atomic weight equation:

(205.98) (x) + (206.98) (x) + (207.98) (1 − 2x) = 207.19

Algebra ensues.

x = 0.263

2) The percent abundances are as follows:

Pb-206: 26.3%
Pb-207: 26.3%
Pb-208: 47.4%

This last form of the question actually falls closer to the real percent abundances of lead than the first form of the question.

Problem #8: If you select one hundred carbon atoms at random, what would the total mass be?

a) 1200.00 amu
b) slightly more than 1200.00 amu
c) slightly less than 1200.00 amu
d) 1300.34 amu
e) slightly less than 1300.34 amu

Solution: Free bauhaus font for mac.

The relevant facts to answer this question are:

There are two naturally-occuring isotopes of carbon: C-12 (12.0 amu) and C-13 (13.0 amu).
The abundance of C-12 is very high, well above 90%.

Let us assume all 100 carbon atoms are C-12. The total weight would be:

12.0 amu x 100 = 1200 amu

Let us now replace a few, say 5, C-12 atoms with C-13. The total weight of the 100 atoms would be:

(12) (95) + (13) (5) = 1205 amu

The correct answer is (b), slightly more than 1200

Problem #9: The atomic weight of naturally occuring neon is 20.18 amu. Naturally occuring neon is composed of two isotopes:

Ne-20: 19.99 amu
Ne-22: 21.99 amu

Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occuring neon.

Solution:

1) Calculate the percent abundance of the two isotopes:

(19.99) (1-x) + (21.99) (x) = 20.18

x = 0.095 (this is the Ne-22)

2) Calculate moles of Ne in 12.55 g:

12.55 g / 20.18 g/mol = 0.6219 mol

3) Calculate atoms of Ne in 0.6219 mol:

0.6219 mol times 6.022 x 10²³ mol^-1 = 3.745 x 10²³

4) Calculate number of Ne-22 atoms:

3.745 x 10²³ times 0.095 = 3.558 x 10²²

Note: the problem above is simplified. There are actually three stable isotopes of neon, with the unmentioned one being Ne-21. Here is a neon problem involving all three isotopes:

The percent abundance for Ne-21 is 0.2700%. Find the percent abundance for Ne-22.

Ne-20 (mass = 19.9924 amu)
Ne-21 (mass = 20.9938 amu)
Ne-22 (mass = 21.9914 amu)

Solution:

Ne-20: abundance = 1 − (x + 0.002700)
Ne-21: abundance = 0.002700
Ne-22: abundance = x

(19.9924) (0.997300 − x) + (20.9938) (0.002700) + (21.9914) (x) = 20.1797

The solution for x is left to the reader.

Note: this problem does not provide the average atomic weight of neon, so it had to be looked up. Also note how 1 minus (x + 0.002700) was rearranged.

Problem #10: Natural rubidium has the average mass 85.4678 amu and is composed of isotopes Rb-85 (mass = 84.9117 amu) and Rb-87. The ratio of atoms Rb-85/Rb-87 in natural rubidium is 2.591. Calculate the mass of Rb-87.

Solution:

Rb-85/Rb-87 is 2.591 to 1. For every 2.591 Rb-85 atoms, there is one Rb-87 atom. We will use 3.591 (coming from 2.591 + 1) to represent 100% of all naturally-occuring Rb atoms.

The percent abundance of Rb-85 is 2.591 / 3.591 = 0.721526

The percent abundance of Rb-87 is 1 / 3.591 = 0.278474 (I actually did it by 1 minus 0.721526)

(84.9117) (0.721526) + (x) (0.278474) = 85.4678

x = 86.9086 amu

Problem #11: Beanium, a new 'element,' has been discovered. Beanium is known to have three naturally-occuring stable 'isotopes.' A sample of 77 'atoms' of beanium was tested and the following results were obtained:

Isotope	Number in Sample	Mass
Black	55	10.26 g
Brown	15	8.64 g
White	7	8.67 g

Calculate the 'atomic weight' of beanium.

Solution:

1) Calculate the percent abundance:

Bl: 55/77 = 0.7143
Br: 15/77 = 0.1948
W: 7/77 = 0.0909

2) Let's get the masses of one of each bean:

Bl: 10.26 / 55 = 0.1865 g
Br: 8.64 / 15 = 0.5760 g
W: 8.67 / 7 = 1.2386 g

3) The 'atomic mass' is a weighted average:

(0.1865) (0.7143) + (0.5760) (0.1948) + (1.2386) (0.0909) = 0.358 g

The 'isotopic' mixture of beans (black, brown and white) apparently weighs 0.358 g per bean atom. You can check this by doing this multiplication:

0.358 g times 77

To a reasonable degree of precision, the answer to the above will equal the sum of the masses of the 77 beans involved.

Problem #12: A sample of boron with a relative atomic mass of 10.8 gives a mass spectrum with two peaks, one at m/z = 10, and one at m/z = 11. Calculate the ratio of the heights of two peaks

Solution:

1) Calculate the percent abundance:

(x) (10) + (1 − x) (11) = 10.8

10x + 11 − 11x = 10.8

x = 0.2

1 − x = 0.8

2) Calculate the peak ratio:

The ratio between 11 and 10 is 0.8 to 0.2.

This is a 4:1 ratio. The peak at m/z = 11 is 4, the peak at m/z = 10 is 1.

Problem #13: Rhenium has two naturally occurring isotopes: Re-185 with a natural abundance of 37.40%, and Re-187 with a natural abundance of 62.60%. The sum of the masses of the two isotopes is 371.9087. What are the atomic weights of Re-185 and Re-187?

Solution:

1) Look up the average atomic weight of Rhenium:

186.207 amu

2) Set up the following:

(0.3740)(x) + (0.6260)(371.9087 − x) = 186.207

where 'x' is the atomic weight of Re-185 and '371.9087 − x' is the atomic weight of Re-187.

3) Algebra ensues:

0.3740x + 232.8148 − 0.6260x = 186.207

-0.2520x = -46.6078

x = 184.952 amu (the atomic weight of Re-185)

Atomic Mass And Number Worksheet

4) The atomic weight of Re-187 is: Dragon ball fighter z crack.

371.9087 minus 184.952 = 186.957 amu

Problem #14: Calculate the percentage of each isotope of mass numbers 36, 37 and 39. The atomic weight of the element is 38.60 amu. Isotopes 36 and 37 are found in equal proportions.

Solution:

1) The first relevant equation:

(36 x a) + (37 x a) + (39 x b) = 38.60 amu [1]

The above equation is the formula used to determine the amu. We know that 36 and 37 are present in equal abundances so we assign them the same unknown, that is to say, a.

39 has some other unknown abundance and we assign it the symbol b.

2) The second relevant equation:

2a + b = 1 [2]

We know that the sum of the abundances must equal 1 (the decimal equivalent for 100%). This is how we obtain equation [2].

3) We now need to solve equation [1] for a and b, but that equation has two unknowns. We use equation [2] to reduce the number of unknowns:

2a + b = 1

b = 1 − 2a [3]

4) We can now substitute equation [3] into [1] and solve:

(36 x a) + (37 x a) + (39 x b) = 38.60 amu

(36 x a) + (37 x a) + (39 x (1 − 2a)) = 38.60 amu

a = 0.08

5) Now substitute value of a into [3]:

b = 1 − 2a
b = 1 − 2(0.08)
b = 0.84

The precent abundances for 36 and 37 are 8% each and the percent abundance for 39 is 84%.

Problem #15: Chromium (atomic mass = 51.9961 amu) has four isotopes. Their masses are 49.94605 amu, 51.94051 amu, 52.94065 amu, and 53.93888 amu. The first two isotopes have a combined abundance of 87.87%, and the last isotope has an abundance of 2.365%. (a) What is the abundance of the third isotope? (b) Estimate the abundance of the first 2 isotopes.

Solution:

1) First question:

If chromium has four isotopes, they have to add up for a sum of 100%, so the abundance of the third one is 9.765%.

Pb Element Atomic Mass

2) Second question:

Let the abundance of the 49.94605 amu isotope be equal to x

Therefore the abundance of the 51.94051 amu is equal to 0.8787 − x (Notice how I am using decimal values, not percentages).

Equation to determine x is:

(x) (49.94605) + (0.8787 − x) (51.94051) + (0.09765) (52.94065) + (0.002365) (53.93888) = 51.9961

Once x is known, 0.8787 − x is easily calculated.

Problem #16: Antimony, Sb, has two stable isotopes. Given that 43.8% of natural antimony is Sb-123 with the experimentally determined mass of 122.904 amu, what is the other stable isotope?

Solution:

The atomic weight of Sb from the periodic table is 121.760. Therefore, the 'weighted average' of the abundance and mass of each of the two isotopes will be:

(0.438) (122.904) + (0.562) (x) = 121.760

53.832 + 0.562x = 121.760

0.562x = 67.928

x = 120.868

The other stable isotope is Sb-121

Atomic Number and Mass Number

When you study the periodic table, the first thing that you may notice is the number that lies above the symbol. This number is known as the atomic number, which identifies the number of protons in the nucleus of ALL atoms in a given element. The symbol for the atomic number is designated with the letter Z. For example, the atomic number (z) for sodium (Na) is 11. That means that all sodium atoms have 11 protons. If you change the atomic number to 12, you are no longer dealing with sodium atoms, but magnesium atoms. Hence, the atomic number defines the element in question.

Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. Because different isotopes of the same element haves different number of neutrons, each of these isotopes will have a different mass number(A), which is the sum of the number of protons and the number of neutrons in the nucleus of an atom.

Mass Number(A) = Number of Protons + Number of Neutrons

The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as ^A_ZX, where X is the atomic symbol of the element, A is the mass number and Z is the atomic number. The isotope of carbon that has 6 neutrons is therefore ¹²₆C. The subscript indicating the atomic number is actually redundant because the atomic symbol already uniquely specifies Z. Consequently, it is more often written as ¹²C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z.

Figure 2.12: Formalism used for identifying specific nuclide (any particular kind of nucleus)

Atomic Mass Unit

The atomic mass unit (u or amu) is a relative unit based on a carbon-12 atom with six protons and six neutrons, which is assigned an exact value of 12 amu's (u's). This is the standard unit for atomic or molecular mass, and 1 amu is thus 1/12^th the mass of a ¹²C atom. This is obviously very small

1 amu = 1.66054x10^-27Kg = 1.66054x10^-24​​​​g

As a result of this standard, the mass of all other elements on the periodic table are determined relative to carbon-12. For example, a Nitrogen-14 atom with 7 protons and 7 neutrons has been experimentally determined to have a mass that is 1.1672 times that of carbon-12. So, the mass of the Nitrogen-14 atom must be 14.00643 u's. As you may have imagined, if another element has been chosen as the standard, the masses of the elements would have been completely different!

Isotopic Distributions

Although carbon-12 is the most abundant type of isotope in carbon, it is not the only isotope. In addition to ¹²C, a typical sample of carbon contains 1.11% ¹³C, with 7 neutrons and 6 protons, and a trace ¹⁴C, with 8 neutrons and 6 protons. The nucleus of ¹⁴C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Table 2.1.3. (Note: we will discuss the derivation of the atomic mass in the next section).

Table 2.1.3: Properties of Selected Isotopes

Element	Symbol	Atomic Mass (amu)	Isotope Mass Number	Isotope Masses (amu)	Percent Abundances (%)
hydrogen	H	1.0079	1	1.007825	99.9855
			2	2.014102	0.0115
boron	B	10.81	10	10.012937	19.91
			11	11.009305	80.09
carbon	C	12.011	12	12 (defined)	99.89
			13	13.003355	1.11
oxygen	O	15.9994	16	15.994915	99.757
			17	16.999132	0.0378
			18	17.999161	0.205
iron	Fe	55.845	54	53.939611	5.82
			56	55.934938	91.66
			57	56.935394	2.19
			58	57.933276	0.33
uranium	U	238.03	234	234.040952	0.0054
			235	235.043930	0.7204
			238	238.050788	99.274

Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991.

Example 2.2.1

An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes.

Given: number of protons and neutrons

Asked for: element and atomic symbol

Strategy:

1. Refer to the periodic table and use the number of protons to identify the element.
2. Calculate the mass number of each isotope by adding together the numbers of protons and neutrons.
3. Give the symbol of each isotope with the mass number as the superscript and the number of protons as the subscript, both written to the left of the symbol of the element.

Solution:

A The element with 82 protons (atomic number of 82) is lead: Pb.

B For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are ²⁰⁶₈₂Pb, ²⁰⁷₈₂Pb, and ²⁰⁸₈₂Pb, which also can also be symbolized as Pb-206, Pb-207, and Pb-208.

Exercise (PageIndex{1})

What is the mass number of a phosphorous atom with 16 neutrons

Answer

31

Exercise (PageIndex{2})

How many protons, neutrons, and electrons are in As-74 atom?

Answer

33 protons, 41 neutrons, 33 electrons

Exercise (PageIndex{3})

The actual mass of As-74 is 73.924 amu's.

A) What is the mass of As-74 in grams?

B) What is the mass of the arsenic atom relative to the carbon-12 atom?

Answer

A) Mass of As-74 = (73.924 amu) x (1.66054 x 10^-24 g/amu) = 1.2275 x 10^-22 g

B) Mass of As-74/Mass of C-12 = 73.924 amu/12 amu= 6.1603 .. so, the mass of an As-74 atom is about six times more than a C-12 atom.

Contributors

• Anonymous
• Modified by Joshua Halpern, Scott Sinex and Scott Johnson
• Bob Belford (UALR) and November Palmer (UALR)
• Ronia Kattoum(UALR)