Question: The table below shows the normal boiling points of several substances. Write a program that prompts the user for the observed boiling point of a substance in Celsius degrees and identifies the substance if the observed boiling point is within 5% of the expected boiling point. If the data input is more than 5% higher or lower than any of the boiling points in the table, the program should output the message “Substance unknown.” Your program should define and call a function “within_x_percent” that takes as parameters a reference value ref, a data value data, and a percentage value x and returns 1 meaning true if data is within x percent of ref –that is,
(ref – x% * ref) < data < (ref + x% * ref). Otherwise “within_x_percent would return zero, meaning false. For example, the call “within_x_percent(357, 323, 10) would return true, since 10% of 357 is 35.7, and 323 falls between 321.3 and 392.7.
Table:
Substance / Normal boiling point (celcius)
Water / 100
Mercury / 357
Copper / 1187
Silver / 2193
Gold / 2660
I keep getting this error:Code:#include <stdio.h> #define WATER_BPT 100 #define MERCURY_BPT 357 #define COPPER_BPT 1187 #define GOLD_BPT 2660 #define SILVER_BPT 2193 int within_x_percent(double ref, double data, double x); int main(void) { double data, x; printf("Please enter the boiling point> "); scanf("lf", &data); x = .05; if(within_x_percent(WATER_BPT, data, x)) printf("The substance is water."); else if(within_x_percent(MERCURY_BPT, data, x)) printf("The substance is mercury."); else if(within_x_percent(COPPER_BPT, data, x)) printf("The substance is copper."); else if(within_x_percent(SILVER_BPT, data, x)) printf("The substance is silver."); else if(within_x_percent(GOLD_BPT, data, x)) printf("The substance is gold."); else printf("The substance is unknown."); return(0); } int within_x_percent(double ref, double data, double x) { return(if(ref-x*ref<=data && data<=ref+x*ref)); }
homework2.c: In function âwithin_x_percentâ:
homework2.c:33:8: error: expected expression before âifâ
Thanks for any help!"



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