All Of The Following Are Equal To Avogadro's Number Except



Avogadro's number isn't a mathematically derived unit. The number of particles in a mole of a material is determined experimentally. This method uses electrochemistry to make the determination. You may wish to review the working of electrochemical cells before attempting this experiment.

All of the following are equal to Avogadro's number EXCEPT O 6.02 x 1023 moles of H2 gas O 22.4 liters of H2 gas 6.02 x 1023 molecules of H2 gas O 2.02 grams of H2 gas.


All of the following are equal to Avogadro's number EXCEPT. The number of atoms of bromine in 1 mol Br b. The number of atoms of gold in 1 mol Au. The statement which is not equalt ot Avogadro's number is the number of atoms of bromine in 1 mol Br₂, as one mole of a substance is equal to 6.022 × 10²³ units of that substance. All of the following are equal to Avogadro's number EXCEPT. The number of atoms of bromine in 1 mol Br₂ b. The number of atoms of gold in 1 mol Au c. The number of molecules of nitrogen in 1 mol N₂ d. The number of molecules of carbon monoxide in 1 mol CO. Avogadro proposed that equal volumes of gases under the same conditions contain the same number of molecules, a hypothesis that proved useful in determining atomic and molecular weights and which led to the concept of the mole. (See Avogadro’s law.) The number of atoms or other particles in a mole is the same for all substances. All of the following are equal to Avogadro's number EXCEPT. The number of atoms of bromine in 1 mol Br b. The number of atoms of gold in 1 mol Au c. The number of molecules of nitrogen in 1 mol N d. The number of molecules of carbon monoxide in 1 mol CO.

The objective is to make an experimental measurement of Avogadro's number.
A mole can be defined as the gram formula mass of a substance or the atomic mass of an element in grams. In this experiment, electron flow (amperage or current) and time are measured in order to obtain the number of electrons passing through the electrochemical cell. The number of atoms in a weighed sample is related to electron flow to calculate Avogadro's number.
In this electrolytic cell both electrodes are copper and the electrolyte is 0.5 M H2SO4. During electrolysis, the copper electrode (anode) connected to the positive pin of the power supply loses mass as the copper atoms are converted to copper ions. The loss of mass may be visible as pitting of the surface of the metal electrode. Also, the copper ions pass into the water solution and tint it blue. At the other electrode (cathode), hydrogen gas is liberated at the surface through the reduction of hydrogen ions in the aqueous sulfuric acid solution. The reaction is:
2 H+(aq) + 2 electrons -> H2(g)
This experiment is based on the mass loss of the copper anode, but it is also possible to collect the hydrogen gas that is evolved and use it to calculate Avogadro's number.
  • Direct current source (battery or power supply)
  • Insulated wires and possibly alligator clips to connect the cells
  • 2 Electrodes (e.g., strips of copper, nickel, zinc, or iron)
  • 250-ml beaker of 0.5 M H2SO4 (sulfuric acid)
  • Water
  • Alcohol (e.g., methanol or isopropyl alcohol)
  • Small beaker of 6 M HNO3 (nitric acid)
  • Ammeter or multimeter
  • Stopwatch
  • Analytical balance capable of measuring to nearest 0.0001 gram
Obtain two copper electrodes. Clean the electrode to be used as the anode by immersing it in 6 M HNO3in a fume hood for 2-3 seconds. Remove the electrode promptly or the acid will destroy it. Do not touch the electrode with your fingers. Rinse the electrode with clean tap water. Next, dip the electrode into a beaker of alcohol. Place the electrode onto a paper towel. When the electrode is dry, weigh it on an analytical balance to the nearest 0.0001 gram.
The apparatus looks superficially like this diagram of an electrolytic cell except that you are using two beakers connected by an ammeter rather than having the electrodes together in a solution. Take beaker with 0.5 M H2SO4 (corrosive!) and place an electrode in each beaker. Before making any connections be sure the power supply is off and unplugged (or connect the battery last). The power supply is connected to the ammeter in series with the electrodes. The positive pole of the power supply is connected to the anode. The negative pin of the ammeter is connected to the anode (or place the pin in the solution if you are concerned about the change in mass from an alligator clip scratching the copper). The cathode is connected to the positive pin of the ammeter. Finally, the cathode of the electrolytic cell is connected to the negative post of the battery or power supply. Remember, the mass of the anode will begin to change as soon as you turn the power on, so have your stopwatch ready!
You need accurate current and time measurements. The amperage should be recorded at one minute (60 sec) intervals. Be aware that the amperage may vary over the course of the experiment due to changes in the electrolyte solution, temperature, and position of the electrodes. The amperage used in the calculation should be an average of all readings. Allow the current to flow for a minimum of 1020 seconds (17.00 minutes). Measure the time to the nearest second or fraction of a second. After 1020 seconds (or longer) turn off the power supply record the last amperage value and the time.
Now you retrieve the anode from the cell, dry it as before by immersing it in alcohol and allowing it to dry on a paper towel, and weigh it. If you wipe the anode you will remove copper from the surface and invalidate your work!
If you can, repeat the experiment using the same electrodes.
Anode mass lost: 0.3554 grams (g)
Current(average): 0.601 amperes (amp)
Time of electrolysis: 1802 seconds (s)
Remember:
one ampere = 1 coulomb/second or one amp.s = 1 coul
charge of one electron is 1.602 x 10-19 coulomb
  1. Find the total charge passed through the circuit.
    (0.601 amp)(1 coul/1amp-s)(1802 s) = 1083 coul
  2. Calculate the number of electrons in the electrolysis.
    (1083 coul)(1 electron/1.6022 x 1019coul) = 6.759 x 1021 electrons
  3. Determine the number of copper atoms lost from the anode.
    The electrolysis process consumes two electrons per copper ion formed. Thus, the number of copper (II) ions formed is half the number of electrons.
    Number of Cu2+ ions = ½ number of electrons measured
    Number of Cu2+ ions = (6.752 x 1021 electrons)(1 Cu2+ / 2 electrons)
    Number of Cu2+ ions = 3.380 x 1021 Cu2+ ions
  4. Calculate the number of copper ions per gram of copper from the number of copper ions above and the mass of copper ions produced.
    The mass of the copper ions produced is equal to the mass loss of the anode. (The mass of the electrons is so small as to be negligible, so the mass of the copper (II) ions is the same as the mass of copper atoms.)
    mass loss of electrode = mass of Cu2+ ions = 0.3554 g
    3.380 x 1021 Cu2+ ions / 0.3544g = 9.510 x 1021 Cu2+ ions/g = 9.510 x 1021 Cu atoms/g
  5. Calculate the number of copper atoms in a mole of copper, 63.546 grams.
    Cu atoms/mole of Cu = (9.510 x 1021 copper atoms/g copper)(63.546 g/mole copper)
    Cu atoms/mole of Cu = 6.040 x 1023 copper atoms/mole of copper
    This is the student's measured value of Avogaro's number!
  6. Calculate percent error.
    Absolute error: |6.02 x 1023 - 6.04 x 1023 | = 2 x 1021
    Percent error: (2 x 10 21 / 6.02 x 10 23)(100) = 0.3 %

All Of The Following Are Equal To Avogadro's Number Excepter Except


Mass is a basic physical property of matter. The mass of an atom or a molecule is referred to as the atomic mass. The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems.

Introduction

In chemistry, there are many different concepts of mass. It is often assumed that atomic mass is the mass of an atom indicated in unified atomic mass units(u). However, the book Quantities, Units and Symbols in Physical Chemistry published by the IUPAC clearly states:

'Neither the name of the physical quantity, nor the symbol used to denote it, should imply a particular choice of unit.'

The name 'atomic mass' is used for historical reasons, and originates from the fact that chemistry was the first science to investigate the same physical objects on macroscopic and microscopic levels. In addition, the situation is rendered more complicated by the isotopic distribution. On the macroscopic level, most mass measurements of pure substances refer to a mixture of isotopes. This means that from a physical stand point, these mixtures are not pure. For example, the macroscopic mass of oxygen (O2) does not correspond to the microscopic mass of O2. The former usually implies a certain isotopic distribution, whereas the latter usually refers to the most common isotope (16O2). Note that the former is now often referred to as the 'molecular weight' or 'atomic weight'.

Mass Concepts in Chemistry
name in chemistryphysical meaningsymbolunits
atomic massmass on microscopic scalem, maDa, u, kg, g
molecular massmass of a moleculemDa, u, kg, g
isotopic massmass of a specific isotopeDa, u, kg, g
mass of entitymass of a chemical formulam, mfDa, u, kg, g
average massaverage mass of a isotopic distribution mDa, u, kg, g
molar massaverage mass per mol M = m/nkg/mol or g/mol
atomic weightaverage mass of an elementAr = m / muunitless
molecular weightaverage mass of a moleculeMr = m / muunitless
relative atomic massratio of mass m and and the atomic mass constant muAr = m / muunitless
atomic mass constantmu = m(12C)/12mu = 1 Da = 1 uDa, u, kg, g
relative molecular massratio of mass m of a molecule and and the atomic mass constant muMr = m / muunitless
relative molar mass???
mass numbernucleon numberAnucleons, or unitless
integer massnucleon number * DamDa, u
nominal massinteger mass of molecule consisting of most abundant isotopesmDa, u
exact massmass of molecule calculated from the mass of its isotopes (in contrast of measured ba a mass spectrometer)Da, u, kg, g
accurate massmass (not normal mass)Da, u, kg, g

These concepts are further explained below.

Average Mass

Isotopes are atoms with the same atomic number, but different mass numbers. A different mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12, by definition). This explains how lithium can have an atomic mass of 6.941 Da. The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature; more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:

[ m(E) = sum_{n=1} m(I_n) times p(I_n) ]

where ∑ represents a n-times summation over all isotopes (I_n) of element E, and p(I) represents the relative abundance of the isotope I.

Example 1

Find the average atomic mass of boron using the Table 1 below:

Mass and abundance of Boron isotopes

n isotope Inmass m (Da)isotopic abundance p
110B10.0130.199
211B11.0090.801

Solution

The average mass of Boron is:

[ m(B) = (10.013 Da)(0.199) + (11.009 Da)(0.801) = 1.99 Da + 8.82 Da = 10.81 Da ]

Relative Mass

Traditionally it was common practice in chemistry to avoid using any units when indicating atomic masses (e.g. masses on microscopic scale). Even today, it is common to hear a chemist say, '12C has exactly mass 12'. However, because mass is not a dimensionless quantity, it is clear that a mass indication needs a unit. Chemists have tried to rationalize the omission of a unit; the result is the concept of relative mass, which strictly speaking is not even a mass but a ratio of two masses. Rather than using a unit, these chemists claim to indicate the ratio of the mass they want to indicate and the atomic mass constant mu which is defined analogous to the unit they want to avoid. Hence the relative atomic mass of the mass m is defined as:

[A_r = dfrac{m}{m_u} ]

The quantity is now dimensionless. As this unit is confusing and against the standards of modern metrology, the use of relative mass is discouraged.

Molecular Weight, Atomic Weight, Weight vs. Mass

Until recently, the concept of mass was not clearly distinguished from the concept of weight. In colloquial language this is still the case. Many people indicate their 'weight' when they actually mean their mass. Mass is a fundamental property of objects, whereas weight is a force. Weight is the force F exerted on a mass m by a gravitational field. The exact definition of the weight is controversial. The weight of a person is different on ground than on a plane. Strictly speaking, weight even changes with location on earth.

When discussing atoms and molecules, the mass of a molecule is often referred to as the 'molecular weight'. There is no univerally-accepted definition of this term; however, mosts chemists agree that it means an average mass, and many consider it dimensionless. This would make 'molecular weight' a synonym to 'average relative mass'.

Integer Mass

Because the proton and the neutron have similar mass, and the electron has a very small mass compared to the former, most molecules have a mass that is close to an integer value when measured in daltons. Therefore it is quite common to only indicate the integer mass of molecules. Integer mass is only meaningful when using dalton (or u) units.

Accurate Mass

Many mass spectrometers can determine the mass of a molecule with accuracy exceeding that of the integer mass. This measurement is therefore called the accurate mass of the molecule. Isotopes (and hence molecules) have atomic masses that are not integer masses due to a mass defect caused by binding energy in the nucleus.

All

Units

The atomic mass is usually measured in the units unified atomic mass unit (u), or dalton (Da). Both units are derived from the carbon-12 isotope, as 12 u is the exact atomic mass of that isotope. So 1 u is 1/12 of the mass of a carbon-12 isotope:

1 u = 1 Da = m(12C)/12

The first scientists to measure atomic mass were John Dalton (between 1803 and 1805) and Jons Jacoband Berzelius (between 1808 and 1826). Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known was Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen. Berzelius demonstrated that this is not always the case by showing that chlorine (Cl) has a mass of 35.45, which is not a whole number multiple of hydrogen's mass.

Some chemists use the atomic mass unit (amu). The amu was defined differently by physicists and by chemists:

  • Physics: 1 amu = m(16O)/16
  • Chemistry: 1 amu = m(O)/16

Chemists used oxygen in the naturally occurring isotopic distribution as the reference. Because the isotopic distribution in nature can change, this definition is a moving target. Therefore, both communities agreed to the compromise of using m(12C)/12 as the new unit, naming it the 'unified atomic mass unit' (u). Hence, the amu is no longer in use; those who still use it do so with the definition of the u in mind. For this reason, the dalton (Da) is increasingly recommended as the accurate mass unit.

Neither u nor Da are SI units, but both are recognized by the SI.

Molar Mass

The molar mass is the mass of one mole of substance, whether the substance is an element or a compound. A mole of substance is equal to Avogadro's number (6.023×1023) of that substance. The molar mass has units of g/mol or kg/mol. When using the unit g/mol, the numerical value of the molar mass of a molecule is the same as its average mass in daltons:

  • Average mass of C: 12.011 Da
  • Molar mass of C: 12.011 g/mol

This allows for a smooth transition from the microscopic world, where mass is measured in daltons, to the macroscopic world, where mass is measured in kilograms.

Example 2

What is the molar mass of phenol, C6H5OH?

Average mass m = 6 × 12.011 Da + 6 × 1.008 Da + 1 × 15.999 Da = 94.113 Da

Molar mass = 94.113 g/mol = 0.094113 kg/mo

Measuring Masses in the Atomic Scale

Masses of atoms and molecules are measured by mass spectrometry. Mass spectrometry is a technique that measures the mass-to-charge ratio (m/q) of ions. It requires that all molecules and atoms to be measured be ionized. The ions are then separated in a mass analyzer according to their mass-to-charge ratio. The charge of the measured ion can then be determined, because it is a multiple of the elementary charge. The the ion's mass can be deduced. The average masses indicated in the periodic table are then calculated using the isotopic abundances, as explained above.

The masses of all isotopes have been measured with very high accuracy. Therefore, it is much simpler and more accurate to calculate the mass of a molecule of interest as a sum of its isotopes than measuring it with a commercial mass spectrometer.

Note that the same is not true on the nucleon scale. The mass of an isotope cannot be calculated accurately as the sum of its particles (given in the table below); this would ignore the mass defect caused by the binding energy of the nucleons, which is significant.

Table 2: Mass of three sub-atomic particles
ParticleSI (kg)Atomic (Da)Mass Number A
Proton1.6726×10-271.00731
Neutron1.6749×10-271.00871
Electron9.1094×10-310.000548580

As shown in Table 2, the mass of an electron is relatively small; it contributes less than 1/1000 to the overall mass of the atom.

Where to Find Atomic Mass

The atomic mass found on the Periodic Table (below the element's name) is the average atomic mass. For example, for Lithium:

The red arrow indicates the atomic mass of lithium. As shown in Table 2 above and mathematically explained below, the masses of a protons and neutrons are about 1u. This, however, does not explain why lithium has an atomic mass of 6.941 Da where 6 Da is expected. This is true for all elements on the periodic table. The atomic mass for lithium is actually the average atomic mass of its isotopes. This is discussed further in the next section.

One particularly useful way of writing an isotope is as follows:

Applications

Applications Include:

All Of The Following Are Equal To Avogadro's Number Except One

  1. Average Molecular Mass
  2. Stoichiometry

Note: One particularly important relationship is illustrated by the fact that an atomic mass unit is equal to 1.66 × 10-24 g. This is the reciprocal of Avogadro's constant, and it is no coincidence:

[dfrac{rm Atomic~Mass~(g)}{1 {rm g}} times dfrac{1 {rm mol}}{6.022 times 10^{23}} = dfrac{rm Mass~(g)}{1 {rm atom}}]

Because a mol can also be expressed as gram × atoms,

[1 u = frac{M_u (molar mass unit)}{N_A (Avogadro's Number)} = 1 frac{g}{mol N_A} ]

1u = Mu(molar mass unit)/NA(Avogadro's Number)=1g/mol/NA

NA known as Avogadro's number (Avogadro's constant) is equal to 6.023×1023 atoms.

Atomic mass is particularly important when dealing with stoichiometry.

Practice Problems

All Of The Following Are Equal To Avogadro's Number Except

  1. What is the molecular mass of radium bicarbonate, Ra(HCO3)2?
  2. List the following, from least to greatest, in terms of their number of neutrons, and then atomic mass: 14N, 42Cl, 25Na, 10Be
  3. A new element, Zenium, has 3 isotopes, 59Ze, 61Ze, and 67Ze, with abundances of 62%, 27%, and 11% respectively. What is the atomic mass of Zenium?
  4. An isotope with a mass number of 55 has 5 more neutrons than protons. What element is it?
  5. How much mass does 3.71 moles of Fluorine have?
  6. How many grams are there in 4.3 × 1022 molecules of POCl3?
  7. How many moles are there in 23 grams of sodium carbonate?

Solutions

a) Molecular mass of Ra(HCO3)2

= 226 + 2(1.01 u + 12.01 u + (16.00 u)(3)) = 348 u or g/mol

b) Number of neutrons: 10Be, 14N, 25Na, 42Cl

Atomic Mass: 10Be, 14N, 25Na, 42Cl

Note: It is the same increasing order for both number of neutrons and atomic mass because more neutrons means more mass.

c) Atomic mass of Zenium:

(59 u)(0.62) + (61 u)(0.27) + (67 u)(0.11)

= 37 u + 16 u + 7.4 u

= 60.4 u or g/mol

d) Mn

e) (3.71 moles F2)(19 × 2 g/mol F2)

= (3.71 mol F2)(38 g/mol F2)

= 140 g F2

f) (4.3 × 1022 molecules POCI3)(1 mol/6.022 × 1023 molecules POCI3)(30.97 + 16.00 + 35.45 x 3 g/mol POCI3)

= (4.3 × 1022 molecules POCI3)(mol/6.022 × 1023 molecules POCI3)(153.32 g/mol POCI3)

= 11 g POCI3

g) (23 g Na2CO3)(1 mol/22.99 × 2 + 12.01 + 16.00 × 3 g Na2CO3)

= (23 g Na2CO3)(1 mol/105.99 g Na2CO3)

= (0.22 mol Na2CO3)

References

  1. Petrucci, Ralph, William Harwood, Geoffrey Herring, and Jeffry Madura. General Chemistry. 9th ed. Upper Saddle River, New Jersey: Pearson Prentice Hall, 2007
  2. Clifford C. Houk, Richard Post. Chemistry Concepts and Problems, a Self-Teaching Guide. 2nd ed. New York : Wiley, 1996.
  3. David R. Lide. CRC Handbook of Chemistry and Physics. 87th ed. New York: CRC Press, 2006.
  4. Loss, R.D., Report of the IUPAC Commission on Atomic Weights and Isotopic Abundances, Chemistry International, 23, 179, 2001.
  5. Mascetta, Joseph A. Chemistry The Easy Way. 3rd ed. New York: Barron's Educational Series, 1996.

Contributors and Attributions

  • Gunitika Dandona (UCD)