Question #9b322

1 Answer
Apr 25, 2017

Answer:

Empirical Formula #=> CO#

Explanation:

#"Sample Mass " - " Carbon Mass = Oxygen Mass"#

so

#"95.2g " - " 40.8g C = 54.4g Oxy"#

Calculate #%# per 100 wt

#%C = (40.8/95.2)100% "w/w" = "42.86% w/w"#

#%O = (100 - 42.86)% = "57.14% w/w"#

Apply Sequence:

#% ("per 100 Wt") => "grams" => "moles" => "mole ratio" => "reduce (divide by smallest mole value)" => "scale to whole no ratio" => "Emp Ratio" => "Emp Formula"#

#%C = 42.86% => "42.86g" => (42.86/12) "mole" => "3.571 mole"#

#%O = 57.14% => "57.14g" => (57.14/16)"mole" => "3.571 mole"#

#"C:O Ratio" => (3.571/3.571):(3.571/3.571) => 1:1# Empirical Ratio #=># Empirical Formula #=> CO#