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Mole factors

Normality is the number of equivalence per liter of solution. It is denoted as N. Normality and molarity are related as follows (a is called the mole factor)... [Pg.51]

Calculation of the composition of a mixture on a mole percentage basis requires the use of mole factors, M. These are obtained by dividing the weight factors by the molar masses of each component of the standard solution and normalizing the resulting numbers. A sample calculation utilizing mole correction factors is provided below in the analysis of a mixture of ethanol, heptane, benzene, and ethyl acetate with a GLC equipped with a thermal conductivity detector. The last column shows the percentage composition calculated directly from the measured peak areas, without correction for detector response. The dramatic differences in the calculated composition with and without this correction, as noted in the last two columns, underscore the importance of this correction for quantitative analysis. [Pg.205]

Analysis Use the quantitative method recommended by your instructor to determine the areas under the two peaks in your gas chromatogram of the test mixture. Divide the weight fraction and mole fraction of each of the two components by the area under the peak for that component. Normalize the value for the weight and mole factors for the unknown by the corresponding weight and mole factors for the standard, and record these values in your notebook. [Pg.208]

Given a quantity in moles of reactant or product, use a mole-mole factor from the balanced equation to calculate the number of moles of another substance In the reaction. [Pg.280]

Mole-Mole Factors from a Balanced Equation... [Pg.282]

From the balanced equation, we see that 2 mol of iron reacts with 3 mol of sulfur to form 1 mol of iron(in) sulfide. Actually, any amount of iron or sulfur may be used, but the ratio of iron reacting with sulfur will always be the same. From the coefficients, we can write mole-mole factors between reactants and between reactants and products. The coefficients used in the mole-mole factors are exact numbers they do not limit the number of significant figures. [Pg.282]

Fe(s) + 3S (5 ) --> Fe2S3(s ), we will use those mole-mole factors in a chemical... [Pg.282]

Step 3 Use coefficients to write mole-mole factors. [Pg.284]

Write all of the mole-mole factors for each of the equations in Problem 9.1. [Pg.284]

For the equations in Problem 9.2, write the setup with the correct mole-mole factor ... [Pg.284]

When we have the balanced chemical equation for a reaction, we can use the mass of one of the substances (A) in the reaction to calculate the mass of another substance (B) in the reaction. However, the calculations require us to convert the mass of A to moles of A using the molar mass factor for A. Then we use the mole-mole factor that links substance A to substance B, which we obtain from the coefficients in the balanced equation. This mole-mole factor (B/A) will convert the moles of A to moles of B. Then the molar mass factor of B is used to calculate the grams of substance B. [Pg.285]

Step 3 Write the mole-mole factors from the equation. [Pg.289]

Step 3 Write the molar mass factors and mole-mole factors from the equation. Molar mass factors... [Pg.291]

The coefficients in an equation describing the relationship between the moles of any two components are used to write mole-mole factors. [Pg.298]

When the number of moles for one substance is known, a mole-mole factor is used to find the moles of a different substance in the reaction. [Pg.298]

In calculations using equations, the molar masses of the substances and their mole-mole factors are used to change the number of grams of one substance to the corresponding grams of a different substance. [Pg.298]

From the coefficients, mole-mole factors can be written for any two substances as follows ... [Pg.300]

Use the mole-mole factor that converts moles of A to moles of B. [Pg.300]

Substitute values into the gas law equation and calculate. When we substitute in the values, we see that the mole factor is greater than 1, which increases volume as predicted in Step 1. [Pg.372]

St6p 3 Write the equalities and conversion factors for molar mass and mole-mole factors. [Pg.379]

Mole-mole factor moles of HCI Molarity liters of HCI solution... [Pg.414]

Step 2 Write a plan to calculate the needed quantity or concentration. We start the problem with the volume and molarity of the HNO3 solution to calculate moles. Then we can use the mole-mole factor and the molar volume to calculate the liters of NO2 gas. [Pg.416]


See other pages where Mole factors is mentioned: [Pg.257]    [Pg.51]    [Pg.282]    [Pg.282]    [Pg.283]    [Pg.285]    [Pg.285]    [Pg.285]    [Pg.291]    [Pg.293]    [Pg.300]    [Pg.300]    [Pg.372]    [Pg.378]    [Pg.379]    [Pg.393]   
See also in sourсe #XX -- [ Pg.205 ]




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Connection between Stoichiometric Factor and Mole Percent Excess Reagent

Conversion factor mass-mole-number relationships

Conversion factor mole

Conversion factor mole ratio

Mole-mass conversion factors

Mole-volume conversion factors

Volume mole-mass conversion factors

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