Empirical formula mass

In the problem above, we determined the percentage data from the chemical formula. We can determine the empirical formula if we know the percent compositions of the various elements. The empirical formula tells us what elements are present in the compound and the simplest whole-number ratio of elements. The data may be in terms of percentage, or mass or even moles. However, the procedure is still the same—convert each element to moles, divide each by the smallest, and then use an appropriate multiplier if necessary. We can then determine the empirical formula mass. If we know the actual molecular mass, dividing the molecular formula mass by the empirical formula mass, gives an integer (rounded if needed) that we can multiply each of the subscripts in the empirical formula. This gives the molecular (actual) formula, which tells what elements are in the compound and the actual number of each. 

The empirical formula mass is 46.00 g/mol. Dividing the actual molar mass by the empirical molar mass gives ... 

C—The general formula, C H2b, means that 1 mol of H20 will form per mole of empirical formula unit, regardless of the value of n. The moles of water formed is the mass of the alkene divided by the empirical formula mass. 

Thus, 24.6 years must be two half-lives of the empirical formula mass. 

Na202. First, find the empirical formula mass of NaO, which is 38.99 g/mol. You determine this by adding one Na (22.99) to one O (16.00). Then divide the grcim formula mass of the mystery compound, 78 g/mol, by this empirical formula mass to obtain the quotient, 2. Multiply each of the subscripts within the empirical formula by this number to obtain Na202. You ve just found the molecular formula for sodium peroxide. 

The empirical formula mass is 97.1 g/mol, which you calculate by multiplying the number of atoms of each element in the compound by the element s atomic mass and adding them all up ... 

Dividing the gram molecular mass you were given (194.2 g/mol) by this empirical formula mass yields the quotient, 2. Multiplying each of the subscripts in the empirical formula by 2 produces the molecular formula, CgHjgN 02. The common name for this culturally important compound is caffeine. 

The solid-state hexamers (2)—(4) at first appeared to dissolve intact in benzene (94). Cryoscopic rmm measurements over a range of concentrations (0.03-0.09 M, molarity expressed relative to the empirical formula mass) implied n values of 5.9-6.1. Furthermore, their room-temperature 7Li NMR spectra in c/8-toluene each consisted of broad singlets within the narrow chemical shift (6) range of + 0.6 to -0.2 ppm (relative to external phenyllithium in the same solvent). However, variations in temperature and concentration affected the 7Li NMR spectra of (2) and, in particular, of (4) (95). Figure 18a shows these spectra for three d8-toluene solutions of (4) at -100°C. The most concentrated solution has a dominant signal at 8 -+0.7, though five or six other signals (indicated by asterisks) are apparent. On dilution,... 

To find the multiple, calculate the ratio of the molecular mass to the empirical formula mass ... 

Elemental analysis can provide only an empirical formula. To determine the molecular formula, it is also necessary to know the substance s molecular mass. In the present problem, the molecular mass of naphthalene is 128.2 amu, or twice the empirical formula mass of C5H4 (64.1 amu). Thus, the molecular formula of naphthalene is C(2x5)H(2x4)/ or Ci0H8. 

The empirical formula mass (88.0 amu) is half the molecular mass of ascorbic acid (176 amu), so the subscripts in the empirical formula must be multiplied by 2 ... 

The empirical formula of caproic acid is therefore C3H6O, and the empirical formula mass is 58.1 amu. Since the molecular mass of caproic acid is 116.2, or twice the empirical formula mass, the molecular formula of caproic acid must be... 

Calculate the empirical formula mass of each of the following ... 

The empirical formula is C3H4. The empirical formula mass is... 

The empirical formula is CH3, with an empirical formula mass of 15.0 g/mol of empirical formula units ... 

The empirical formula is CH2. The empirical formula mass is therefore 14.0 amu per empirical formula unit. The number of empirical formula units per molecule is given by 28.1 amu/molecule 14.0 amu/empirical formula unit... 

Divide the molecular mass by the empirical formula mass. The result must be an integer. 

A compound consists of 92.26% C and 7.74% H. Its molecular mass is 65.0 amu. (a) Calculate its empirical formula, (b) Calculate its empirical formula mass, (c) Calculate the number of empirical formula units in one molecule, (d) Calculate its molecular formula. 

Define or identify each of the following molecule, ion, formula unit, formula mass, mole, molecular mass, Avogadro s number, percent, empirical formula, molecular formula, molar mass, empirical formula mass, molecular weight. 

Dividing both numbers of moles by 7.40 yields 1.00 mol C and 1.50 mol H. Multiplying both of these by 2 yields the empirical formula C2H3. The empirical formula mass is thus 27.0 amu. The number of empirical formula units in 1 mol can be calculated by using 110 amu for the molecular mass. The number must be an integer. 

Suppose we know that this compound with empirical formula CH5N has a molar mass of 31.06. How do we determine which of the possible choices represents the molecular formula Since the molecular formula is always a whole-number multiple of the empirical formula, we must first find the empirical formula mass for CH5N ... 

To obtain the molecular formula, we must compare the empirical formula mass with the molar mass. The empirical formula mass for P205 is 141.94. 

In Example 3.6 we found the molecular formula by comparing the empirical formula mass with the molar mass. There is an alternative way to obtain the molecular formula. The molar mass and the percentages (by mass) of each element present can be used to compute the moles of each element present in one mole of compound. These numbers of moles then represent directly the subscripts in the molecular formula. This procedure is illustrated in Example 3.7. 

Calculate the empirical formula mass using the molar mass of each element. 

Now that we know that the empirical formula for the compound is CH, we determine the mass of the empirical formula to be 13.0 u (12.0 u + 1.01 u = 13.0 u). The problem stated that the molecular mass was 26.0 u, which is double the empirical formula mass, so we multiply each subscript by 2 to get the molecular formula of 2(CH) = C2H2. 

From the analytical data, an empirical formula of CHCl is calculated. The empirical formula mass is 48.47 g- mol". The problem may be solved using the ideal gas law ... 

The n can be calculated from adipic acid s molecular mass and its empirical formula mass. A substance s empirical formula mass can be calculated from the subscripts in its empirical formula and the atomic masses of the elements. The empirical formula mass of adipic acid is... 

Because the subscripts in the molecular formula are always a positive integer multiple of the subscripts in the empirical formula, the molecular mass is always equal to a positive integer multiple of the empirical formula mass. 

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