The steps in determining empirical formula

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Presentation transcript:

The steps in determining empirical formula

The steps in performing this process Step One- Divide by the mass Step Two- Divide by the lowest numbers of mols Step Three- Decimal Formed After Simplifying Multiplier Whole Number Formed .5 2 1 .25 4 .75 3 .33 .99 or 1 .66 1.98 or 2

Multiplier is 2 based on the 2.5 Analysis of a 10.150g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of a 4.433g. What is the empirical formula of this compound. (10.150-4.433=5.717 gr) Step #1 4.433 gr P 5.717 gr O 30.974 15.999 Step #2 .143 mols .3573 mols .143 .143 Step #3 1 atom P 2.4986 atom O (X 2) P2O5 (X 2) Multiplier is 2 based on the 2.5

Step #3 2 atom Na 1 atom S 3 atoms O A compound is found to contain 36.48% Na, 25.41% S, and 38.11% O. Find its empirical formula. Step #1 36.48 gr Na 25.41 gr S 38.11 gr O 22.9897 32.066 15.999 Step #2 1.5861 mols .7941 mols 2.3879 mols .7941 .7941 .7941 Step #3 2 atom Na 1 atom S 3 atoms O Na2SO3 There are no multipliers

Multiplier is 2 based on the 1.501 Find the empirical formula of a compound that contains 53.70% iron and 46.30% sulfur. Step #1 53.70 gr Fe 46.30 gr S 55.845 32.066 Step #2 .96159 mols 1.4439 mols .96159 .96159 Step #3 1 atom Fe 1.501 atom S (X 2) Fe2S3 (X 2) Multiplier is 2 based on the 1.501

No multipliers!!!!!!!!!! Step #1 8.83 gr Na 6.17 gr S 22.9897 32.066 A 15.0g sample of a compound is found to contain 8.83g sodium and 6.17g sulfur. Calculate the empirical formula of this compound. Step #1 8.83 gr Na 6.17 gr S 22.9897 32.066 Step #2 .3841 mols .1924 mols .1924 .1924 Step #3 2 atoms Na 1 atom S Na2S No multipliers!!!!!!!!!!

FeS There is no multiplier Step #1 63.52 gr Fe 36.48 gr S A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula. Step #1 63.52 gr Fe 36.48 gr S 55.845 32.066 Step #2 1.1374 mols 1.1377 mols 1.1374 1.1374 Step #3 1 atom Fe 1.0003 atom S FeS There is no multiplier

Step #3 1.989 atom Na 1 atom S 3.97 atoms O Qualitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. Step #1 32.38 gr Na 22.65 gr S 44.99 gr O 22.9897 32.066 15.999 Step #2 1.4078 mols .7078 mols 2.812 mols .7078 .7078 .7078 Step #3 1.989 atom Na 1 atom S 3.97 atoms O Na2SO4 There are no multipliers

Step #3 1 atom K 1 atom Cr 3.49 atoms O Find the empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium and the remainder oxygen. (100-26.56-35.41=38.03% O) Step #1 26.56 gr K 35.74 gr Cr 38.03 gr O 39.0983 51.996 15.999 Step #2 .681 mols .681 mols 2.377 mols .681 .681 .681 Step #3 1 atom K 1 atom Cr 3.49 atoms O K2Cr2O7 The multiplier for 3.5 is 2. K-1X2, Cr 1X2 and Cr 3.5 X2

A 60.00g sample of tetraethyl lead, a gasoline additive, is found to contain 38.43g lead, 17.83g carbon, and 3.74g hydrogen. Find its empirical formula. Step #1 38.43 gr Pb 17.83 gr C 3.74 gr H 207.2 12.011 1.008 Step #2 .1855 mols 1.4858 mols 3.710 mols .1855 .1855 .1855 Step #3 1 atom Pb 8.0097 atom C 20 atoms H PbC8H20

CaBr2 There is no multiplier Step #1 4.0 gr Ca 16.00 gr Br Analysis of a 20.0g sample of a compound containing only calcium and bromine indicates that 4.00g of calcium are present. What is the empirical formula of the compound formed? (20.0-4.00=16.0 gr Br). Step #1 4.0 gr Ca 16.00 gr Br 40.078 79.904 Step #2 .098087 mols .20024 mols .098087 .098087 Step #3 1 atom Ca 2.0415 atom Br CaBr2 There is no multiplier

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