Verify whether glucose (C6H12O6) fits the empirical formula (CH2O)n. Identify n and state the atomic ratio of C:H:O for glucose.

n = 6; atomic ratio C:H:O = 6:12:6 which simplifies to 1:2:1.

n = 2; atomic ratio C:H:O = 6:12:6 simplified to a unit of C3H6O3 giving n = 2 for that reduced formula representation.

n = 12; atomic ratio C:H:O = 6:12:6 but because hydrogen count is doubled, you must double the unit to CH4O2 and get n = 12 to balance empirical subscripts.

n = 3; atomic ratio C:H:O = 6:12:6 but treated as 3 repeating units of C2H4O2 so the empirical n for glucose is 3 when simplified by mass considerations.