A question about gram atomic, gram molecular, and gram formula mass...?

Question

I know that gram atomic mass deals with elements' atomic mass expressed in grams, and that this number in grams is also how much 1 mole of that element weighs. So my question is, does gram molecular mass and gram formula mass tell you how much 1 mole of that molecule or complex molecule weighs?

Thanks :)

Answer 1

HELLO how are YOU!!

The sum of the atomic weights of all atoms making up a molecule. Actually, what is meant by molecular weight is molecular mass. The use of this expression is historical, however, and will be maintained. The atomic weight is the mass, in atomic mass units, of an atom. It is approximately equal to the total number of nucleons, protons and neutrons composing the nucleus. Since 1961 the official definition of the atomic mass unit (amu) has been that it is 1/12 the mass of the carbon-12 isotope, which is assigned the value 12.000 exactly. See also Atomic mass unit; Atomic mass unit; Relative atomic mass; Relative molecular mass.

A mole is an amount of substance containing the Avogadro number, NA, approximately 6.022 × 1023, of molecules or atoms. Molecule, in this definition, is understood to be the smallest unit making up the characteristic compound. Originally, the mole was interpreted as that number of particles whose total mass in grams was numerically equivalent to the atomic or molecular weight in atomic mass units, referred to as gram-atomic or gram-molecular weight. This is how the above value for NA was calculated. As the ability to make measurements of the absolute masses of single atoms and molecules has improved, however, modern metrology is tending to alter its approach and define the Avogadro number as an exact quantity, thereby changing slightly the definition of the atomic mass unit and removing the need to define atomic weight with respect to a particular isotopic species. The latest and most accurate value for the Avogadro number is 6.0221415(10) × 1023 mol?1. See also Avogadro number; Mole (chemistry).

As the masses of all the atomic species are now well known, masses of molecules can be determined once the composition of the molecule has been ascertained. Alternatively, if the molecular weight of the molecule is known and enough additional information about composition is available, such as the basic atomic constituents, it is possible to begin to assemble structural information about the molecule. Thus, the determination of the molecular weight is one of the first steps in the analysis of an unknown species. Given the increasing emphasis on the study of biologically important molecules, particular attention has been focused on the determination of molecular weights of larger and larger units. There are a number of methods available, and the one chosen will depend on the size and physical state of the molecule. All processes are physical macroscopic measurements and determine the molecular weight directly. Connection to the absolute mass scale is straightforward by using the Avogadro number, although, for extremely large molecules, this connection is often unnecessary or impossible, as the accuracy of the measurements is not that good. The main function of molecular weight determination of large molecules is elucidation of structure.

Molecular weight determination of materials which are solid or liquid at room temperature is best achieved by taking advantage of one of the colligative properties of solutions, boiling-point elevation, freezing-point lowering, or osmotic pressure, which depend on the number of particles in solution, not on the nature of the particle. The choice of which to use will depend on a number of properties of the substance, the most important of which will be the size. All require that the molecule be small enough to dissolve in the solution but large enough not to participate in the phase change or pass through a semipermeable membrane. Freezing-point lowering is an excellent method for determining molecular weights of smaller organic molecules, and osmometry, as the osmotic pressure determination is called, for determining molecular weights of larger organic molecules, particularly polymeric species. Boiling-point elevation is used less frequently. See also Polymer.

The basis of all the methods involving colligative properties of solutions is that the chemical potentials of all phases must be the same. (Chemical potential is the partial change in energy of a system as matter is transferred into or out of it. For two systems in contact at equilibrium, the chemical potentials for each must be equal.) See also Chemical equilibrium; Chemical thermodynamics.

Another measurement from which molecular weights can be obtained is based on the scattering of light from the molecule. A beam of light falling on a molecule will induce in the molecule a dipole moment which in its turn will radiate. The interference between the radiated beam and the incoming beam produces an angular dependence of the scattered radiation which depends on the molecular weight of the molecule. This occurs whether the molecule is free or in solution. While the theory for this effect is complicated and varies according to the size of the molecule, the general result for molecules whose size is considerably less than that of the wavelength ? of the radiation (less than ?/50) is given by the equation below; I(θ) is the intensity of radiation at angle θ, I0 the intensity of the incoming beam, M the molecular weight, and c the concentration in grams per cubic centimeter of the molecule. If the molecules are much larger than ?/50 (about 9 nanometers for visible light), this relationship in this simple form is no longer valid, but the method is still viable with appropriate adjustments to the theory. In fact, it can be used in its extended version even for large aggregates.

Answer 2

Yes you take the sum of the masses of the elemnts of the molecule.
Do not forget to multiply each mass of atom by their number in the molecule

Ex H2SO4
in a periodic table you find atomic mass H= 1 , S= 32 O =16

So molecular mass 1*2 + 32 + 16*4 = 98 g

Answer 3

Yup, you've got it the only thing to be careful of is how accurate your teacher will want the significant figures.