To calculate the percent error in this case, you need to compare the experimental value (the calculated molecular mass) to the accepted value (the value given by the teacher). Here's the correct calculation:
Experimental value: 959.4 g/mol
Accepted value: 122.2 g/mol
Percent error = ((Experimental value - Accepted value) / Accepted value) * 100
Percent error = ((959.4 - 122.2) / 122.2) * 100
Percent error = (837.2 / 122.2) * 100
Percent error = 685.4%
So the correct percent error is 685.4%. It seems like there was a slight error in your calculation.