The listed atomic mass represents an average atomic mass of naturally occurring carbon. It was calculated as follows: average atomic mass of C = (.9890)(12 amu) + (.01100)(13.00335) = 12.01 amu. No real carbon atom has a mass of 12.01 amu. Most carbon atoms have an atomic mass of 12 amu and a few have an atomic mass of 13.00335 amu. Symbol: C Atomic Number: 6 Atomic Mass: 12.0107 amu Melting Point: 3500.0 °C (3773.15 K, 6332.0 °F) Boiling Point: 4827.0 °C (5100.15 K, 8720.6 °F) Number of Protons/Electrons: 6 Number of Neutrons: 6 Classification: Non-metal Crystal Structure: Hexagonal Density @ 293 K: 2.62 g/cm 3 Color: May be black Atomic Structure.

• Relative Atomic Mass Of C
• Atomic Mass Of Carbon Dioxide
• Atomic Mass Of S
• Atomic Mass Of Calcium
• Atomic Mass Of Cm

Lecture 3

Atomic Mass

Covalent compounds are substances formed when non-metallic atoms combine chemically.Most compounds have a fixed atomic composition. The formula of a substance specifies its atomic composition.For example, glucose, a simple sugar, has the formula C₆H₁₂O₆.This means that 6 carbon atoms are bonded to 12 hydrogen atoms and 6 oxygen atoms, in each molecule of glucose.In theory, we could combine elemental C, O and H in the correct proportions and make a molecule of glucose.Carbon, in its most common state, exists as the solid, graphite.Oxygen exists as the diatomic gas molecule, O₂ and hydrogen exists as the diatomic gas, H₂.

The smallest detectable speck of graphite would contain about 1 x 10¹⁶ atoms of carbon, so we are not going to be able to weigh a single atom.What has been done, instead, is to assign a value to the mass of one carbon atom.By definition, the atomic mass of ¹²C (carbon with 6 protons and 6 neutrons) has been set as 12 amu (atomic mass units).However, when you look up the atomic mass of carbon in a table, it is shown as 12.01 amu.This is because carbon exists as a number of different isotopes.The most common isotopes of carbon are ¹²C and ¹³C (6 protons and 7 neutrons).The amu of ¹³C has been determined to be 13.00335 amu.It is heavier than ¹²C because it has an extra neutron and because neutrons have slightly higher masses than protons.¹²C is more abundant (98.9%) than ¹³C (1.10%), which is why the listed mass of C is closer to 12 (12.01) than it is to 13.00335.The listed atomic mass represents an average atomic mass of naturally occurring carbon.It was calculated as follows:

average atomic mass of C = (.9890)(12 amu) + (.01100)(13.00335) = 12.01 amu

No real carbon atom has a mass of 12.01 amu.Most carbon atoms have an atomic mass of 12 amu and a few have an atomic mass of 13.00335 amu.The atomic mass of H has been determined to be 1.0079 amu and that of O to be 15.9994 amu.Again, these are average values of the common isotopes of these elements.

Avogadro's Number and Molar Mass

Now we have a measure of the relative masses of the elements.It would be of practical use to somehow convert amu to grams, a unit that we can actually measure.Even the smallest sample of an element contains an enormous number of atoms.Therefore, it would be convenient to have a special unit, which describes a very large number of atoms.This unit is called a mole and is the SI unit of quantity. Paper is sold by reams (500 sheets), eggs are sold by the dozen (12) atoms and molecules are measured in moles.A mole is defined as the number of ¹²C atoms in exactly 12 grams of ¹²C.This number has been determined, experimentally, to be 6.022 x 10²³ and is called Avogadro's number, N_A.

6.022 x 10²³ atoms ¹²C = 12 g ¹²C

1 atom ¹²C = 12 amu

6.022 x 10²³ atoms = 1 mol atoms

With these conversion factors, we can determine the mass of a single ¹²C atom.

12 g ¹²C6.022 x 10²³ atoms= 12 g/mol = molar mass of ¹²C

6.022 x 10²³ atoms1 mol

The atomic masses for all of the elements have been tabulated as amu.This same number is also the molar mass, in g/mol, of each element.

So, if we weighed out a 12.01 g sample of carbon, we would have 6.022 x 10²³ atoms of carbon in our sample.Looking at our formula for glucose (C₆H₁₂O₆) we can see that we need twice as many H atoms as C atoms.Our 12.01 g sample of carbon contains 1 mole of carbons atoms.We will, therefore, need 2 moles of H atoms.How many grams of H atoms would this be?We can use the conversion factor relating amu to grams/mol of a substance.

2 mol H1.0079 g H= 2.0158 g H

mol H

How much oxygen would be needed?The formula indicates that for every mole of carbon we need one mole of oxygen.So, we will need 1 mol of O atoms.

1 mol O15.9994 g O= 15.9994 g O

mol O

Our starting materials would weight 12.01g + 2.0158g + 15.9994 g = 30.03 g.

This would also be the mass of the glucose that could be made from these elements.How many moles and molecules of glucose would this be?

The mass of one molecule of glucose, C₆H₁₂O₆, would be the sum of the atomic masses of its elements:

6 x 12.011 (amu of C) + 12 x 1.0079 (amu of H) + 6 x 15.9994 (amu of O) = 180.157 amu

180.157 amu = 180.157 g/mol

30.03 g glucose1 molglucose= 0.1667 moles of glucose

180.157 g glucose

0.1667 moles glucose6.022 x 10²³ molecules glucose = 1.004 x 10²³ molecules of glucose

mol glucose

Percent Composition

Another way of describing the composition of a substance is in terms of its percent composition, the percentage, by mass,of the elements in that substance.This is information that is experimentally obtainable, which can be used to derive the empirical formula of a compound, as will be discussed below.For now, lets define what is meant by percent composition by determining the percent composition of sodium nitrite, NaNO₂.First, calculate the molar mass:

molar mass =1 mol Na (22.99 g/mol)

+ 1 mol N(14.01 g/mol)

+ 2 mol O(16.00 g/mol)

69.00 g/mol NaNO₃

The percent composition of the elements is then:

% Na =22.99 g x 100% = 33.32%

69.00 g

% N =14.01 g x 100% = 20.30%

69.00 g

%O =32.00 g x 100% = 46.38%

69.00 g

It is possible to use this method to determine the formula of an unknown substance.

Analysis of an unknown compound shows the following percent composition:

40.92 % carbon, by weight

Relative Atomic Mass Of C

4.58 % hydrogen, by weight

54.50% oxygen, by weight

First, assume that you are dealing with some amount of the unknown, say, 100 grams.

40.92 % = 0.4092 x 100 g = 40.92 g of C

4.58 % = 0.058 x 100 g = 4.58 g of H

54.50 % = 0.5450 x 100 g = 54.50 g O

Since atoms combine on a molar basis, not by masses, convert these grams of elements to moles of the different elements.

40.95 g C1 mol C= 3.407 mol C

12.01 g C

4.58 g H1 mol H= 4.54 mol H

1.008 g H

54.5 g O1 mol O= 3.406 mol O

16.00 g O

These numbers indicate the relative number of moles of each of the three elements in the compound.We can now write a formula based on them:

C_3.407 H_4.54 O_3.40

However, whole atoms combine to form molecules, not fractional atoms.So, divide each of these factors by the smallest factor, 3.406.This gives:

CH_1.333O

There is still a fractional subscript.Find a factor, which will convert 1.333 to a whole number:

1.333 x 1 = 1.333

1.333 x 2 = 2.666

1.333 x 3 = 4.000

1.333 x 4 = 5.333

Now multiply all of the subscripts by this factor:

C₃H₄O₃

Atomic Mass Of Carbon Dioxide

This is called the empirical formula, which tells us the relative numbers of each type of atom in this molecule.This means that the molecule could be:

C₃H₄O₃

C₆H₈O₆

C₉H₁₂O₉

in other words, (C₃H₄O₃)_n

The empirical formula mass (C₃H₄O₃) is:3(12.01 g/mol C)

Atomic Mass Of S

+ 4(1.008 g/mol H)

+ 3(16.00 g/mol O)

88.06 g/mol

This means that the molecular mass will be some multiple of this value.If we are told that the molecular mass is 176.12 g/mol, we can determine the molecular formula.

(C₃H₄O₃)_n = 176.12 g/mol

(C₃H₄O₃)= 88.06 g/mol

(88.06)_n = 176.12

n = 2

Atomic Mass Of Calcium

So, the molecular formula would be C₆H₈O₆.

The most common use of these calculations is to determine the empirical mass for a new or unknown compounds based on the products produced by burning the unknown (combustion reaction).In this reaction, all of the carbon in the compound is converted to CO₂, carbon dioxide.All of the hydrogen in the compound is converted to H₂O, water.The mass of the CO₂ and H₂O are carefully measured, and then used to arrive at an empirical formula.

11.5 g of an unknown compound is burned, producing 22.0 g of CO₂ and 13.5 g of H₂O.What is the empirical formula of the compound?

All of the carbon in the CO₂ came from the sample.So, first calculate the number of moles of carbon in the 22.0 g of CO₂.

22.0 g CO₂mol CO₂1 molC= 0.500 moles C

44.01 g CO₂1 mol CO₂

All of the hydrogen in the water came from the unknown, so calculate the number of moles of hydrogen in the 13.5 g of H₂O.

13.5 g H₂O1 mol H₂O2 mol H= 1.50 moles H

18.03 g H₂O1 mol H₂O

We also have to determine if any of the oxygen in the CO₂ and H₂O came from the unknown, or whether it was environmental oxygen used in the combustion reaction.To determine this, we need to compare the mass of the unknown to the mass of the hydrogen and carbon that we know came from the unknown.

mass_unknown=mass _hydrogen + mass_carbon + mass_oxygen

mass _hydrogen = 1.50 mol H1.0079 g H =1.51 g H

1 mol H

mass_carbon = 0.500 mol O12.011 g C = 6.00 g O

1 mol C

mass_unknown = 11.5 g = 1.51 g + 6.00 g + mass_oxygen

mass_oxygen = 4.0 g

Atomic Mass Of Cm

So, 4.0 g of the oxygen must have come from the unknown.Convert this to moles of oxygen

4.0 g O1 mol O=0.25 moles O

15.9994 g O

Now we can determine the empirical formula of the unknown.First, substitute the calculated numbers of moles into the formula:

C_0.5H_1.5O_0.25

Divide by the smallest fraction (0.25), since these are non-integer subscripts.This gives:

C₂H₆O

In order to determine the molecular formula, we would need to know the molar mass to determine if the unknown is C₂H₆O or C₄H₁₂O₂ or C₆H₁₈O₃.

In the modern periodic table, the elements are listed in order of increasing atomic number. The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 6 protons is a carbon atom, no matter how many neutrons may be present). The number of protons determines how many electrons surround the nucleus, and it is the arrangement of these electrons that determines most of the chemical behavior of an element.

In a periodic table arranged in order of increasing atomic number, elements having similar chemical properties naturally line up in the same column (group). For instance, all of the elements in Group 1A are relatively soft metals, react violently with water, and form 1+ charges; all of the elements in Group 8A are unreactive, monatomic gases at room temperature, etc. In other words, there is a periodic repetition of the properties of the chemical elements with increasing mass.

In the original periodic table published by Dimitri Mendeleev in 1869, the elements were arranged according to increasing atomic mass— at that time, the nucleus had not yet been discovered, and there was no understanding at all of the interior structure of the atom, so atomic mass was the only guide to use. Once the structure of the nucleus was understood, it became clear that it was the atomic number that governed the properties of the elements.